Indefinite Integrals

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Exercise 19.2

Question 1.

Evaluate the following integrals:

Answer:

Given:

By Splitting, we get,

By using the formula,

⇒ + +

Question 2.

Evaluate the following integrals:

Answer:

Given:

By Splitting them, we get,

By using the formula,

By using the formula,

⇒

By using the formula,

Question 3.

Evaluate the following integrals:

Answer:

Given:

By Splitting, we get,

By using the formula

Question 4.

Evaluate the following integrals:

∫ (2 – 3x)(3 + 2x)(1 – 2x)dx

Answer:

Given:

⇒∫(2 - 3x)(3 + 2x)(1 - 2x)dx

By multiplying,

⇒∫ (6 - 4x - 9x - 6x²) dx

⇒∫ (6 - 13x - 6x²) dx

By Splitting, we get,

⇒∫6dx -∫13 x dx -∫6x² dx

By using the formulas,

We get,

Question 5.

Evaluate the following integrals:

Answer:

Given:

By Splitting, we get,

By using formula,

By using the formula,

By using the formula,

Question 6.

Evaluate the following integrals:

Answer:

Given:

By applying (a - b)² = a² - 2ab + b²

After computing,

By Splitting, we get,

By applying the formulas:

We get,

I = 1/2 x² + logx - 2x + c

Question 7.

Evaluate the following integrals:

Answer:

Given:

Applying: (a + b)³ = a³ + b³ + 3ab² + 3a²b

⇒

⇒

By Splitting, we get,

By applying formula,

Question 8.

Evaluate the following integrals:

Answer:

Given:

By Splitting, we get,

By applying formula,

⇒

Question 9.

Evaluate the following integrals:

∫ (x^e + e^x + e^e) dx

Answer:

Given:

By Splitting, we get,

By using the formula,

By applying the formula,

We know that,

Question 10.

Evaluate the following integrals:

Answer:

Given:

Opening the bracket, we get,

By multiplying,

By applying the formula,

Question 11.

Evaluate the following integrals:

Answer:

Given:

By multiplying with inside brackets,

By Splitting them, we get,

By applying the formula,

Question 12.

Evaluate the following integrals:

Answer:

Given:

By applying: a³ + b³ = (a + b)(a² + b² - ab)

By Splitting

By using the formula,

Question 13.

Evaluate the following integrals:

Answer:

Given:

By Splitting them,

⇒+

By applying the formula,

We get,

Question 14.

Evaluate the following integrals:

Answer:

Given:

By applying (a + b)² = a² + b² + 2ab

By Splitting, we get,

Question 15.

Evaluate the following integrals:

∫√x(3 – 5x) dx

Answer:

Given:

By multiplying √x inside the bracket we get,

⇒

By Splitting, we get,

By using the formula,

Question 16.

Evaluate the following integrals:

Answer:

Given:

By Splitting,

By applying the formula,

Question 17.

Evaluate the following integrals:

Answer:

Given:

By Splitting, we get,

By applying,

By Splitting, we get,

By applying the formula,

Question 18.

Evaluate the following integrals:

∫ (3x + 4)² dx

Answer:

Given:

By applying,

(a + b)² = a² + b² + 2ab

By Splitting, we get,

By applying,

Question 19.

Evaluate the following integrals:

Answer:

Given:

Take x is common on both numerator and denominator,

Splitting 7x² into 4x² and 3x²

Common the 2x² from first two elements and 3x from next elements,

Now common the from the elements

Now Splitting, we get,

Now applying the formula,

Question 20.

Evaluate the following integrals:

Answer:

Given:

Now spilt 12x³ into 7x³ and 5x³

Now common 5x³ from two elements 7x from other two elements,

Now Splitting, we get,

Question 21.

Evaluate the following integrals:

Answer:

Given:

We know that,

sin²x = 1 – cos²x

We treat 1 – cos²x as a² – b² = (a + b)(a - b)

By Splitting, we get,

We know that,

⇒x – sin x + c

Question 22.

Evaluate the following integrals:

∫ (se²x + cosec²x) dx

Answer:

Given:

By Splitting, we get,

By applying the formula,

⇒tan x – cot x + c

Question 23.

Evaluate the following integrals:

Answer:

Given:

By Splitting, we get,

By cancelling the sin²x on first and cos²x on second,

We know that,

We know that,

⇒secx - (- cotx) + c

⇒secx + cotx + c

Question 24.

Evaluate the following integrals:

Answer:

Given:

By Splitting we get,

We know that,

We know that,

Question 25.

Evaluate the following integrals:

∫ (tan x + cot x)² dx

Answer:

Given:

We know that,

tan²x = sec²x - 1

cot²x = cosec²x - 1

We know that,

I=tanx – cotx - c

Question 26.

Evaluate the following integrals:

Answer:

Let

We know cos2θ = 1 – 2sin²θ = 2cos²θ – 1

Hence, in the numerator, we can write 1 – cos2x = 2sin²x

In the denominator, we can write 1 + cos2x = 2cos²x

Therefore, we can write the integral as

[∵ sec²θ – tan²θ = 1]

Recall and

∴ I = tan x – x + c

Thus,

Question 27.

Evaluate the following integrals:

Answer:

Let

On multiplying and dividing (1 + cos x), we can write the integral as

[∵ sin²θ + cos²θ = 1]

[∵ cosec²θ – cot²θ = 1]

Recall and

We also have

∴ I = –cosec x – cot x – x + c

Thus,

Question 28.

Evaluate the following integrals:

Answer:

Let

We know cos2θ = 2cos²θ – 1 = cos²θ – sin²θ

Hence, in the numerator, we can write cos²x – sin²x = cos2x

In the denominator, we can write 4x = 2 × 2x

⇒ 1 + cos4x = 1 + cos(2×2x)

⇒ 1 + cos4x = 2cos²2x

Therefore, we can write the integral as

Recall

Thus,

Question 29.

Evaluate the following integrals:

Answer:

Let

On multiplying and dividing (1 + cos x), we can write the integral as

[∵ sin²θ + cos²θ = 1]

Recall

We also have

∴ I = –cot x – cosec x + c

Thus,

Question 30.

Evaluate the following integrals:

Answer:

Let

On multiplying and dividing (1 + sin x), we can write the integral as

[∵ sin²θ + cos²θ = 1]

Recall

We also have

∴ I = tan x + sec x + c

Thus,

Question 31.

Evaluate the following integrals:

Answer:

Let

On multiplying and dividing (sec x – tan x), we can write the integral as

[∵ sec²θ – tan²θ = 1]

Recall and

We also have

∴ I = sec x – tan x + x + c

Thus,

Question 32.

Evaluate the following integrals:

Answer:

Let

On multiplying and dividing (cosec x + cot x), we can write the integral as

[∵ cosec²θ – cot²θ = 1]

Recall

We also have

∴ I = –cot x – cosec x + c

Thus,

Question 33.

Evaluate the following integrals:

Answer:

Let

We know cos2θ = 2cos²θ – 1

Hence, in the denominator, we can write 1 + cos2x = 2cos²x

Therefore, we can write the integral as

Recall

Thus,

Question 34.

Evaluate the following integrals:

Answer:

Let

We know cos2θ = 1 – 2sin²θ

Hence, in the denominator, we can write 1 – cos2x = 2sin²x

Therefore, we can write the integral as

Recall

Thus,

Question 35.

Evaluate the following integrals:

Answer:

Let

We know cos2θ = 2cos²θ – 1

Hence, in the denominator, we can write 1 + cos2x = 2cos²x

In the numerator, we have sin2x = 2sinxcosx

Therefore, we can write the integral as

Recall

Thus,

Question 36.

Evaluate the following integrals:

Answer:

Let

We know sinθ = cos(90° - θ)

Therefore, we can write the integral as

Recall and

Thus,

Question 37.

Evaluate the following integrals:

Answer:

Let

We know cos2θ = 1 – 2sin²θ

Hence, in the denominator, we can write 1 – cos2x = 2sin²x

In the numerator, we have sin2x = 2sinxcosx

Therefore, we can write the integral as

Recall

Thus,

Question 38.

Evaluate the following integrals:

Answer:

Let

We know

Therefore, we can write the integral as

Recall

∴ I = x² + c

Thus,

Question 39.

Evaluate the following integrals:

Answer:

Let

We know a³ + b³ = (a + b)(a² – ab + b²)

Hence, in the numerator, we can write

x³ + 8 = x³ + 2³

⇒ x³ + 8 = (x + 2)(x² – x × 2 + 2²)

⇒ x³ + 8 = (x + 2)(x² – 2x + 4)

Therefore, we can write the integral as

Recall and

Thus,

Question 40.

Evaluate the following integrals:

Answer:

Let

We know (a + b)² = a² + 2ab + b²

Therefore, we can write the integral as

We have sec²θ – tan²θ = cosec²θ – cot²θ = 1

Recall and

We also have

⇒ I = a²tan x + b²(–cot x) – (a - b)² × x + c

∴ I = a²tan x – b²cot x – (a – b)²x + c

Thus,

Question 41.

Evaluate the following integrals:

Answer:

Let

Recall and

We also have and

Thus,

Question 42.

Evaluate the following integrals:

Answer:

Let

On multiplying and dividing (1 – cos x), we can write the integral as

[∵ sin²θ + cos²θ = 1]

[∵ cosec²θ – cot²θ = 1]

Recall and

We also have

⇒ I = –cosec x – (–cot x) + x + c

⇒ I = –cosec x + cot x + x + c

Thus,

Question 43.

Evaluate the following integrals:

Answer:

Let

We have

We know cos2θ = 1 – 2sin²θ = 2cos²θ – 1

Hence, in the numerator, we can write

In the denominator, we can write

Therefore, we can write the integral as

[∵ sec²θ – tan²θ = 1]

Recall and

Thus,

Question 44.

Evaluate the following integrals:

Answer:

Let

We have sec²θ – tan²θ = cosec²θ – cot²θ = 1

Recall and

We also have and

⇒ I = 3(–cos x) – 4(sin x) + 6(tan x) – 7(–cot x) + c

∴ I = –3cosx – 4sinx + 6tanx + 7cotx + c

Thus,

Question 45.

If and, find f(x).

Answer:

Given and

On integrating the given equation, we have

We know

Recall

On substituting x = 1 in f(x), we get

On substituting the value of c in f(x), we get

Thus,

Question 46.

If f’(x) = x + b, f(1) = 5, f(2) = 13, find f(x).

Answer:

Given f’(x) = x + b, f(1) = 5 and f(2) = 13

On integrating the given equation, we have

We know

Recall and

On substituting x = 1 in f(x), we get

….. (1)

On substituting x = 2 in f(x), we get

⇒ 13 = 2 + 2b + c

⇒ 13 – 2 = 2b + c

⇒ 2b + c = 11 ….. (2)

By subtracting equation (1) from equation (2), we have

On substituting the value of b in equation (1), we get

∴ c = –2

On substituting the values of b and c in f(x), we get

Thus,

Question 47.

If f’(x) = 8x³ – 2x, f(2) = 8, find f(x).

Answer:

Given f’(x) = 8x³ – 2x and f(2) = 8

On integrating the given equation, we have

We know

Recall

⇒ f(x) = 2x⁴ – x² + c

On substituting x = 2 in f(x), we get

f(2) = 2(2⁴) – 2² + c

⇒ 8 = 32 – 4 + c

⇒ 8 = 28 + c

∴ c = –20

On substituting the value of c in f(x), we get

f(x) = 2x⁴ – x² + (–20)

∴ f(x) = 2x⁴ – x² – 20

Thus, f(x) = 2x⁴ – x² – 20

Question 48.

If f’(x) = a sin x + b cos x and f’(0) = 4, f(0) = 3, , find f(x).

Answer:

Given f’(x) = a sin x + b cos x and f’(0) = 4

On substituting x = 0 in f’(x), we get

f’(0) = asin0 + bcos0

⇒ 4 = a × 0 + b × 1

⇒ 4 = 0 + b

∴ b = 4

Hence, f’(x) = a sin x + 4 cos x

On integrating this equation, we have

We know

Recall and

On substituting x = 0 in f(x), we get

f(0) = –acos0 + 4sin0 + c

⇒ 3 = –a × 1 + 4 × 0 + c

⇒ 3 = –a + c

⇒ c – a = 3 -------------- (1)

On substituting in f(x), we get

⇒ 5 = –a × 0 + 4 × 1 + c

⇒ 5 = 0 + 4 + c

⇒ 5 = 4 + c

∴ c = 1

On substituting c = 1 in equation (1), we get

1 – a = 3

⇒ a = 1 – 3

∴ a = –2

On substituting the values of c and a in f(x), we get

f(x) = –(–2)cos x + 4sinx + 1

∴ f(x) = 2cosx + 4sinx + 1

Thus, f(x) = 2cosx + 4sinx + 1

Question 49.

Write the primitive or anti-derivative of.

Answer:

Given

Let

Recall

Thus, the primitive of f(x) is

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Exercise 19.11

Question 1.

Evaluate the following integrals:

∫ tan³ x sec² x dx

Answer:

Let tan x = t, then

⇒sec² x dx = dt

Question 2.

Evaluate the following integrals:

∫ tan x sec⁴ x dx

Answer:

Let tan x = t, then

⇒sec² x dx = dt

Question 3.

Evaluate the following integrals:

∫ tan⁵ x sec⁴ x dx

Answer:

Let tan x = t, then

⇒sec² x dx = dt

Question 4.

Evaluate the following integrals:

∫ sec⁶ x tan x dx

Answer:

Substituting, sec x = t ⇒ sec x tan x dx = dt

Question 5.

Evaluate the following integrals:

∫ tan⁵ x dx

Answer:

Let tan x = t, then

⇒sec² x dx = dt

Question 6.

Evaluate the following integrals:

Answer:

Let tan x = t, then

⇒sec² x dx = dt

Question 7.

Evaluate the following integrals:

∫ sec⁴ 2x dx

Answer:

Let tan 2x = t, then

⇒2sec² 2x dx = dt

Question 8.

Evaluate the following integrals:

∫ cosec⁴ 3x dx

Answer:

Let cot 3x = t, then

⇒ - 3cosec² 3x dx = dt

Question 9.

Evaluate the following integrals:

∫ cotⁿ x cosec² x dx, n ≠ –1

Answer:

Let cot x = t ⇒ - cosec² x dx = dt

Question 10.

Evaluate the following integrals:

∫ cot⁵ x cosec⁴ x dx

Answer:

Let cot x = t, then

⇒ - cosec² x dx = dt

Question 11.

Evaluate the following integrals:

∫ cot⁵ x dx

Answer:

Let cot x = t, then

⇒ - cosec² x dx = dt

Question 12.

Evaluate the following integrals:

∫ cot⁶ x dx

Answer:

Let cot x = t, then

⇒ - cosec² x dx = dt

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Exercise 19.12

Question 1.

Evaluate the following integrals:

∫ sin⁴ x cos³ x dx

Answer:

Let

We know the Differentiation of

So,

substitute all in above equation,

∫ sin⁴ x cos³ x dx =

=

=

=

=

We know, basic integration formula, ∫xⁿ dx = + c for any c≠-1

Hence,

Put back t = sin x

∫ sin⁴ x cos³ x dx =

Question 2.

Evaluate the following integrals:

∫ sin⁵ x dx

Answer:

∫ sin⁵ x dx = ∫

= ∫ { since sin²x + cos²x = 1}

= ∫

= ∫( sin (

= ∫( sin ( { since sin²x + cos²x = 1}

= ∫( sin

= ∫ sin (separate the integrals)

We know , d(cos x) = -sin xdx

So put cos x = t and dt = -sin xdx in above integrals

= ∫ sin

= ∫ sin

= ∫ sin

= ∫ sin

= ∫ sin

= ( since ∫xⁿ dx = + c )

Put back t = cos x

=

=

= -[

Question 3.

Evaluate the following integrals:

∫ cos⁵ x dx

Answer:

∫ cos⁵ x dx = ∫

= ∫ { since sin²x + cos²x = 1}

= ∫

= ∫( cos (

= ∫( cos ( { since sin²x + cos²x = 1}

= ∫( cos

= ∫ cos (separate the integrals)

We know , d(sin x) = cos xdx

So put sin x = t and dt = cos xdx in above integrals

= ∫ cos

= ∫ cos

= ∫ cos

= ∫ cos

= ∫ cos

= ( since ∫xⁿ dx = + c )

Put back t = sin x

=

Question 4.

Evaluate the following integrals:

∫ sin⁵ x cos x dx

Answer:

Let sin x = t

Then d(sin x) = dt = cos xdx

Put t = sin x and dt = cos xdx in above equation

∫ sin⁵ x cos x dx =

= ( since ∫xⁿ dx = + c for any c≠-1)

=

Question 5.

Evaluate the following integrals:

∫ sin³ x cos⁶ x dx

Answer:

Since power of sin is odd, put cos x = t

Then dt = -sin xdx

Substitute these in above equation,

∫ sin³ x cos⁶ x dx =

=

=

=

= ( since ∫xⁿ dx = + c )

=

Question 6.

Evaluate the following integrals:

∫ cos⁷ x dx

Answer:

∫ cos⁷ x dx =

=

= { since sin²x + cos²x = 1}

We know (a-b)³ = a³b³- 3a²b + 3ab²

Here, a = 1 and b = sin² x

Hence,

= ) {take cos xdx inside brackets)

= (separate the integrals)

Put sinx = t and cos xdx = dt

=

=

=

Put back t = sin x

=

Question 7.

Evaluate the following integrals:

∫ x cos³ x² sin x² dx

Answer:

Let cosx² = t

Then d(cosx²) = dt

Since d(xⁿ) = nxⁿ⁻¹ and d(cos x) = -sinx dx

dt = 2x (-sin x²) = -2x sin x² dx

x sin x²dx =

hence ∫ x cos³ x² sin x² dx =

=

=

=

Question 8.

Evaluate the following integrals:

∫ sin⁷ x dx

Answer:

∫ sin⁷ x dx =

=

= { since sin²x + cos²x = 1}

We know (a-b)³ = a³ - b³ - 3a²b + 3ab²

Here, a = 1 and b = cos² x

Hence,

= ) {take sin xdx inside brackets)

= (separate the integrals)

Put cosx = t and -sinx dx = dt

=

=

=

Put back t = cos x

=

Question 9.

Evaluate the following integrals:

∫ sin³ x cos⁵ x dx

Answer:

Let cos x = t then dt = -sin xdx

Substitute all these in the above equation,

∫ sin³ x cos⁵ x dx =

=

=

=

=

= ( since ∫xⁿ dx = + c )

=

=

Question 10.

Evaluate the following integrals:

Answer:

=

Adding the powers : -4 + -2 = -6

Since all are even nos, we will divide each by cos⁶x to convert into positive power

So, =

=

= { since sec²x = 1 + tan²x}

= ( apply (a + b)² = a² + b² + 2ab )

Let tanx = t, so dt = d(tanx) = sec²xdx

So,

Put t and dx in the above equation,

=

=

=

=

=

=

= {1/tanx = cotx)

Question 11.

Evaluate the following integrals:

Answer:

=

Adding the powers , -3 + -5 = -8

Since it is an even number, we will divide numerator and denominator by cos⁸x

=

= = =

= ∫

We know, (a + b)³ = a³ + b³ + 3a²b + 3ab²

Here, a = 1 and b = tan²x

Hence, ∫ = ∫

Let tan x = t, then dt = d(tanx) = sec²xdx

Put these values in above equation:

= ( since ∫xⁿ dx = + c and ∫t⁻¹ dt = logt)

=

=

Question 12.

Evaluate the following integrals:

Answer:

=

Adding the powers , -3 + -1 = -4

Since it is an even number, we will divide numerator and denominator by cosx

=

= =

= ∫

Let tan x = t, then dt = d(tanx) = sec²xdx

Put these values in the above equation:

= ( since ∫xⁿ dx = + c and ∫t⁻¹ dt = logt)

=

=

Question 13.

Evaluate the following integrals:

Answer:

We know, sin²x + cos²x = 1

Therefore

Divide each term of numerator separately by

= =

= (divide second term each by cos²x )

=

Therefore,

=

=

Put tanx = t, dt = sec²x dx

=

=

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Exercise 19.13

Question 1.

Evaluate the following integrals:

Answer:

PUT x = a sinθ, so dx = a cosθ dθ and θ = sin⁻(x/a)

Above equation becomes,

= = {take a² outside)

=

= = (sec²θ-1 = tan²θ)

= =

=

Put θ = sin⁻(x/a)

=

Question 2.

Evaluate the following integrals:

Answer:

PUT x = a sinθ, so dx = a cosθ dθ and θ = sin⁻(x/a)

Above equation becomes,

= = {take a² outside)

= =

= =

Put θ = sin⁻(x/a)

=

Question 3.

Evaluate the following integrals:

Answer:

and

=

We know 1 + cos 2t = 2cos²t and 1-2cos2t = 2sin²t

Hence,

Therefore , =

Put

=

Question 4.

Evaluate the following integrals:

Answer:

let x = tanθ , so dx = sec²θ dθ and

Putting above values ,

= =

=

Put

=

Question 5.

Evaluate the following integrals:

Answer:

= (add and substract 1)

= ( =

=

Put x + 1 = t hence dx = dt and x = t-1

=

=

We have,

Here a = 3

Therefore,

Put t = x + 1

=

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Exercise 19.14

Question 1.

Evaluate the following integrals:

Answer:

Taking out b²,

=

= { since

=

Question 2.

Evaluate the following integrals:

Answer:

take out a²

=

= { since

=

Question 3.

Evaluate the following integrals:

Answer:

take out a²

=

= { since }

=

Question 4.

Evaluate the following integrals:

Answer:

Add and subtract 4 in the numerator, we get

= =

= {separate the numerator terms)

=

= { since }

=

Question 5.

Evaluate the following integrals:

Answer:

Let I = =

Let t = 2x, then dt = 2dx or dx = dt/2

Therefore,

=

{since

=

Question 6.

Evaluate the following integrals:

Answer:

Let bx = t then dt = bdx or

Hence, =

= {since

Put t = bx

=

Question 7.

Evaluate the following integrals:

Answer:

Let bx = t then dt = bdx or

Hence, =

= {since

Put t = bx

=

Question 8.

Evaluate the following integrals:

Answer:

Let (2-x) = t , then dt = -dx , or dx = -dt

Hence, =

= - {since

Put t = 2-x

=

Question 9.

Evaluate the following integrals:

Answer:

Let (2-x) = t , then dt = -dx , or dx = -dt

Hence, =

= - {since

Put t = 2-x

=

Question 10.

Evaluate the following integrals:

Answer:

We will use basic formula : (a + b)² = a² + b² + 2ab

Or, a² + b² = (a + b)² -2ab

Here,

=

Applying above formula, we get,

x⁴ + 1 = (x² + 1)² - 2 × 1 × x²

=

Hence, =

Separate the numerator terms,

=

{ add and subtract 2 to the second term)

=

{ }

=

=

{ since }

=

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Exercise 19.15

Question 1.

Evaluate the following integrals:

Answer:

let

Let .....(i)

⇒ dx = dt

so,

[using (i)]

Question 2.

Evaluate the following integrals:

Answer:

let

Let .....(i)

⇒ dx = dt

so,

[using (i)]

Question 3.

Evaluate the following integrals:

Answer:

: let

Question 4.

Evaluate the following integrals:

Answer:

let

Let ……(i)

⇒ dx = dt

so,

[using (i)]

Question 5.

Evaluate the following integrals:

Answer:

We have,

=

Sol,

Let x+3 =t

Then dx = dt

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Exercise 19.16

Question 1.

Evaluate the following integrals:

Answer:

let

Let tan x = t .....(i)

⇒ dx = dt

so,

[using (i)]

Question 2.

Evaluate the following integrals:

Answer:

: let

Let = t .....(i)

⇒ dx = dt

so,

[using(i)]

Question 3.

Evaluate the following integrals:

Answer:

Let

Let sin x = t .....(i)

⇒ cos x dx = dt

So,

Again, let t + 2 = u …..(ii)

⇒ dt = du

[using(i),(ii)]

Question 4.

Evaluate the following integrals:

Answer:

let

Let = t .....(i)

⇒ dx = dt

Let .....(i)

⇒ dt = du

so,

[using (i)]

[using (ii)]

Question 5.

Evaluate the following integrals:

Answer:

let

Let = t .....(i)

⇒ dx = dt

[using (i)]

Question 6.

Evaluate the following integrals:

Answer:

let

Let = t .....(i)

⇒ dx = dt

[using (i)]

Question 7.

Evaluate the following integrals:

Answer:

Let

Let = t .....-----(i)

⇒ 2x dx = dt

Put t + 1 = u .....-----(ii)

⇒ dt = du

[using (i)]

[using (ii)]

Question 8.

Evaluate the following integrals:

Answer:

let

Let = t .....(i)

⇒ dx = dt

[using (i)]

Question 9.

Evaluate the following integrals:

Answer:

let

Let = t .....(i)

⇒ dx = dt

[using (i)]

Question 10.

Evaluate the following integrals:

Answer:

let

Let = t .....(i)

⇒ dx = dt

[using (i)]

Question 11.

Evaluate the following integrals:

Answer:

let

Let = t .....(i)

⇒ dx = dt

[using (i)]

[log m – log n= ]

Question 12.

Evaluate the following integrals:

Answer:

Let

Let = t .....-----(i)

⇒ 2x dx = dt

Put t – 1/2 = u .....-----(ii)

⇒ dt = du

[using (i)]

[using (ii)]

Question 13.

Evaluate the following integrals:

Answer:

Let

Let = t .....(i)

⇒ 2x dx = dt

Put t – 3 = u .....(ii)

⇒ dt = du

[using (ii)]

[using (i)]

Question 14.

Evaluate the following integrals:

Answer:

To evaluate the following integral following steps:

Let .....(i)

⇒ dx = dt

Now

= log |(1+t)| - log |(2+t)| + c

[log m – log n= ]

[using(i)]

Question 15.

Evaluate the following integrals:

Answer:

let

Multiply and divide by sin⁡x

Let cot x = t

-cosec x dx =dt

So,

---

Exercise 19.17

Question 1.

Evaluate the following integrals:

Answer:

let

let (x-1)=t

dx=dt

so,

Question 2.

Evaluate the following integrals:

Answer:

8+3x-x2 can be written as 8-

Therefore

Let x-3/2=t

dx=dt

Question 3.

Evaluate the following integrals:

Answer:

Let I =

Let (x + 1) = t

Differentiating both sides, we get,

dx = dt

So,

Question 4.

Evaluate the following integrals:

Answer:

let

dx=dt

Question 5.

Evaluate the following integrals:

Answer:

let

[β>α]

Let (x-(α+β)/2)=t

dx=dt

Question 6.

Evaluate the following integrals:

Answer:

let

dx=dt

Question 7.

Evaluate the following integrals:

Answer:

let

dx=dt

Question 8.

Evaluate the following integrals:

Answer:

7-6x-x² can be written as 7-(x²+6x+9-9)

Therefore

7-(x²+6x+9-9)

Let x+3=t

dx=dt

Question 9.

Evaluate the following integrals:

Answer:

we have

completing the square

Put x-1/5=t then dx =dt

Therefore

---

Exercise 19.18

Question 1.

Evaluate the following integrals:

Answer:

Let x² = t , so 2x dx = dt

Or, x dx =

Hence,

Since,

Hence,

Put t = x²

=

=

Question 2.

Evaluate the following integrals:

Answer:

Let tan x = t

Then

Therefore,

Since,

Hence,

Question 3.

Evaluate the following integrals:

Answer:

Let

Then we have,

Therefore,

Since we have,

Hence,

Question 4.

Evaluate the following integrals:

Answer:

Let

Then

Hence,

Since we have,

Therefore,

=

Question 5.

Evaluate the following integrals:

Answer:

Let

Then

Or,

Therefore, =

Since,

Therefore,

Question 6.

Evaluate the following integrals:

Answer:

Let x² = t

2x dx = dt or x dx = dt/2

Hence,

Since we have,

So,

Put t = x²

=

Question 7.

Evaluate the following integrals:

Answer:

Put 3logx = t

We have

Hence,

Or

Hence,

Since we have,

Hence,

Put t = 3logx

Question 8.

Evaluate the following integrals:

Answer:

Let t = sin²4x

dt = 2sin4x cos4x × 4 dx

we know sin2x = 2sins2xcos2x

therefore, dt = 4 sin8x dx

or, sin8x dx = dt/4

Since we have,

=

Question 9.

Evaluate the following integrals:

Answer:

Let = sin2x

dt = 2cos2x dx

Cos2x dx = dt/2

=

Since we have,

Question 10.

Evaluate the following integrals:

Answer:

Let t = sin²x

dt = 2sinx cosx dx

we know sin2x = 2sins2xcos2x

therefore, dt = sin2x dx

Add and subtract 2² in denominator

Let t + 2 = u

dt = du

Since,

=

Question 11.

Evaluate the following integrals:

Answer:

Let t = cos²x

dt = 2cosx sinx dx = - sin2x dx

therefore,

since, [ sin²x = 1 - cos²x]

Since,

Question 12.

Evaluate the following integrals:

Answer:

Let sinx = t

dt = cosxdx

therefore, =

Since we have,

=

Question 13.

Evaluate the following integrals:

Answer:

Let

So,

Or,

=

Since,

=

Question 14.

Evaluate the following integrals:

Answer:

Let sin^–1x = t

Therefore, =

Since we have,

=

Question 15.

Evaluate the following integrals:

Answer:

Let sinx = t

Cosx dx = dt

Add and subtract 1² in denominator

Let t - 1 = u

dt = du

Since,

Put u = t - 1

Put t = sinx

=

Question 16.

Evaluate the following integrals:

Answer:

Since cosec x = 1/ sinx

Multiply with (1 + sinx) both numerator and denominator

Since (a + b) × (a - b) = a² - b²,

Let sinx = t

dt = cosx dx

therefore,

multiply and divide by 2 and add and subtract (1/2)² in denominator,

Let t + 1/2 = u

dt = du

Since,

Question 17.

Evaluate the following integrals:

Answer:

=

Let sinx + cosx = t

(Cosx - sinx) = dt

Therefore,

Since,

Question 18.

Evaluate the following integrals:

Answer:

= = dx

Let sinx + cosx = t

(Cosx - sinx) = dt

Therefore,

Since we have,

---

Exercise 19.19

Question 1.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for x² + 3x + 2 and I can be reduced to a fundamental integration.

As,

∴ Let, x = A(2x + 3) + B

⇒ x = 2Ax + 3A + B

On comparing both sides –

We have,

2A = 1 ⇒ A = 1/2

3A + B = 0 ⇒ B = –3A = –3/2

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ – I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = x² + 3x + 2 ⇒ du = (2x + 3)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with

∴ I₂ =

⇒ I₂ =

⇒ I₂ = …eqn 3

From eqn 1:

I = I₁ – I₂

Using eqn 2 and eqn 3:

I =

Question 2.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for x² + x + 3 and I can be reduced to a fundamental integration.

As,

∴ Let, x = A(2x + 1) + B

⇒ x = 2Ax + A + B

On comparing both sides –

We have,

2A = 1 ⇒ A = 1/2

A + B = 0 ⇒ B = –A = –1/2

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ – I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As I₁ =

Let u = x² + x + 3 ⇒ du = (2x + 1)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting the value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with

∴ I₂ =

⇒ I₂ = …eqn 3

From eqn 1:

I = I₁ – I₂

Using eqn 2 and eqn 3:

I =

Question 3.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for x² + 2x –4 and I can be reduced to a fundamental integration.

As,

∴ Let, x – 3 = A(2x + 2) + B

⇒ x – 3 = 2Ax + 2A + B

On comparing both sides –

We have,

2A = 1 ⇒ A = 1/2

2A + B = –3 ⇒ B = –3–2A = –4

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ – 4I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = x² + 2x – 4 ⇒ du = (2x + 2)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with

∴ I₂ = …eqn 3

From eqn 1:

I = I₁ – 4I₂

Using eqn 2 and eqn 3:

I =

I =

Question 4.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make a substitution for x² + 6x + 13 and I can be reduced to a fundamental integration.

As

∴ Let, 2x – 3 = A(2x + 6) + B

⇒ 2x – 3 = 2Ax + 6A + B

On comparing both sides –

We have,

2A = 2 ⇒ A = 1

6A + B = –3 ⇒ B = –3–6A = –9

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ – 9I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = x² + 6x + 13 ⇒ du = (2x + 6)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with

∴ I₂ = …eqn 3

From eqn 1:

I = I₁ – 9I₂

Using eqn 2 and eqn 3:

I =

I =

Question 5.

Evaluate the integral:

Answer:

I = =

⇒ I =

Let, sin x = t ⇒ cos x dx = dt

∴ I =

As we can see that there is a term of t in numerator and derivative of t² is also 2t. So there is a chance that we can make substitution for t² – 7t + 12 and I can be reduced to a fundamental integration.

As,

∴ Let, 3t – 2 = A(2t – 7) + B

⇒ 3t – 2 = 2At – 7A + B

On comparing both sides –

We have,

2A = 3 ⇒ A = 3/2

–7A + B = –2 ⇒ B = 7A – 2 = 17/2

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = t² – 7t + 12 ⇒ du = (2t – 7)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

∵ I₂ =

⇒ I₂ =

Using: a² – 2ab + b² = (a – b)²

We have:

I₂ =

I₂ matches with the form

∴ I₂ =

I₂ =

…eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I =

Putting value of t in I:

I = …..ans

Question 6.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for 3x² –4x + 3 and I can be reduced to a fundamental integration.

As,

∴ Let, x – 1 = A(6x – 4) + B

⇒ x – 1 = 6Ax – 4A + B

On comparing both sides –

We have,

6A = 1 ⇒ A = 1/6

–4A + B = –1 ⇒ B = –1+4A = –2/6 = –1/3

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ – I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = 3x² – 4x + 3 ⇒ du = (6x – 4)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in the denominator.

∴ I₂ = {on taking 3 common from denominator}

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with

∴ I₂ =

∴ …eqn 3

From eqn 1:

I = I₁ – I₂

Using eqn 2 and eqn 3:

I =

Question 7.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for 3x² +13x – 10 and I can be reduced to a fundamental integration.

As,

∴ Let, x + 7 = A(6x + 25) + B

⇒ x + 7 = 6Ax + 25A + B

On comparing both sides –

We have,

6A = 1 ⇒ A = 1/6

25A + B = 5 ⇒ B = –25A + 5 = 5/6

Hence,

I =

∴ I =

Let, I₁ =and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = 3x² + 25x +28 ⇒ du = (6x + 25)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ =and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with the form

∴ I₂ =

∴ …eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I =

Question 8.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for –x² + x +2 and I can be reduced to a fundamental integration.

As,

∴ Let, 2x = A(–2x + 1) + B

⇒ 2x = –2Ax + A + B

On comparing both sides –

We have,

–2A = 2 ⇒ A = –1

A + B = 0 ⇒ B = –A = 1

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = 2 + x – x²⇒ du = (–2x + 1)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with

∴ I₂ =

∴ …eqn 3

From eqn 1:

I = I₁ + I₂

Using eqn 2 and eqn 3:

∴ I =

Question 9.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for 3x² + 4x + 2 and I can be reduced to a fundamental integration.

As,

∴ Let, 1–3x = A(6x + 4) + B

⇒ 1–3x = 6Ax + 4A + B

On comparing both sides –

We have,

6A = –3 ⇒ A = –1/2

4A + B = 1 ⇒ B = –4A+1 = 3

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As I₁ =

Let u = 3x² + 4x + 2 ⇒ du = (6x + 4)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting the value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with

∴ I₂ =

∴ …eqn 3

From eqn 1:

I = I₁ + I₂

Using eqn 2 and eqn 3:

∴ I =

Question 10.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for x² – x –2 and I can be reduced to a fundamental integration.

As,

∴ Let, 2x + 5 = A(2x – 1) + B

⇒ 2x + 5= 2Ax – A + B

On comparing both sides –

We have,

2A = 2 ⇒ A = 1

–A + B = 5 ⇒ B = A + 5 = 6

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = x² – x – 2 ⇒ du = (2x – 1)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² – 2ab + b² = (a – b)²

We have:

I₂ =

I₂ matches with

∴ I₂ =

∴ …eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I = …..ans

Question 11.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x³ in numerator and derivative of x⁴ is also 4x³. So there is a chance that we can make substitution for x⁴ + c² and I can be reduced to a fundamental integration but there is also a x term present. So it is better to break this integration.

I = = I₁ + I₂ …eqn 1

I₁ =

As,

To make the substitution, I₁ can be rewritten as

∴ Let, x⁴ + c² = u

⇒ du = 4x³ dx

I₁ is reduced to simple integration after substituting u and du as:

∴ I₁ = …eqn 2

As,

I₂ =

∵ we have derivative of x² in numerator and term of x² in denominator. So we can apply method of substitution here also.

As, I₂ =

Let, x² = v

⇒ dv = 2x dx

∵ I₂ = =

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

I₂ matches with

∴ I₂ =

⇒ I₂ = …eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I = …..ans

Question 12.

Evaluate the integral:

Answer:

I = =

⇒ I =

Let, sin x = t ⇒ cos x dx = dt

∴ I =

As we can see that there is a term of t in numerator and derivative of t² is also 2t. So there is a chance that we can make substitution for t² – 4t + 4 and I can be reduced to a fundamental integration.

As,

∴ Let, 3t – 2 = A(2t – 4) + B

⇒ 3t – 2 = 2At – 4A + B

On comparing both sides –

We have,

2A = 3 ⇒ A = 3/2

–4A + B = –2 ⇒ B = 4A – 2 = 4

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = t² – 4t + 4 ⇒ du = (2t – 4)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ =

I₁ = ….eqn 2

∵ I₂ =

⇒ I₂ =

Using: a² – 2ab + b² = (a – b)²

We have:

I₂ =

As,

∴ I₂ = + C …eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I =

Putting value of t in I:

I = …..ans

Question 13.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for 2x² +6x +5 and I can be reduced to a fundamental integration.

As,

∴ Let, x + 2 = A(4x + 6) + B

⇒ x + 2 = 4Ax + 6A + B

On comparing both sides –

We have,

4A = 1 ⇒ A = 1/4

6A + B = 2 ⇒ B = –6A + 2 = 1/2

Hence,

I =

∴ I =

Let, I₁ =and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = 2x² + 6x + 5 ⇒ du = (4x + 6)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with 1𝑥2+𝑎2𝑑𝑥=112 𝑡𝑎𝑛−132112+ 𝐶I₂ matches with the form

∴ I₂ =

∴ …eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I = …..ans

Question 14.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for 3x² +2x +1 and I can be reduced to a fundamental integration.

As,

∴ Let, 5x – 2 = A(6x + 2) + B

⇒ 5x – 2 = 6Ax + 2A + B

On comparing both sides –

We have,

6A = 5 ⇒ A = 5/6

2A + B = –2 ⇒ B = –2A – 2 =–11/3

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = 3x² + 2x + 1 ⇒ du = (6x + 2)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with the form

∴ I₂ =

∴ …eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I =

Question 15.

Evaluate the integral:

Answer:

I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for 3x² +13x – 10 and I can be reduced to a fundamental integration.

As,

∴ Let, x + 5 = A(6x + 13) + B

⇒ x + 5 = 6Ax + 13A + B

On comparing both sides –

We have,

6A = 1 ⇒ A = 1/6

13A + B = 5 ⇒ B = –13A + 5 =17/6

Hence,

I =

∴ I =

Let, I₁ =and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = 3x² + 13x – 10 ⇒ du = (6x + 13)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with the form

∴ I₂ =

∴ …eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I =

Question 16.

Evaluate the integral:

Answer:

Let, I =

I =

If we assume x² to be an another variable, we can simplify the integral as derivative of x² i.e. x is present in numerator.

Let, x² = u

⇒ 2x dx = du

⇒ x dx = 1/2 du

∴ I =

As,

∴ Let, u = A(2u + 1) + B

⇒ u = 2Au + A + B

On comparing both sides –

We have,

2A = 1 ⇒ A = 1/2

A + B = 0 ⇒ B = –A = –1/2

Hence,

I =

∴ I =

Let, I₁and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let v = u² + u + 1 ⇒ dv = (2u + 1)du

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

∴ I₂ =

∴ …eqn 3

From eqn 1, we have:

I = I₁ + I₂

Using eqn 2 and 3, we get –

I =

Putting value of u in I:

I =

I =

Question 17.

Evaluate the integral:

Answer:

Let, I =

I =

If we assume x² to be an another variable, we can simplify the integral as derivative of x² i.e. x is present in numerator.

Let, x² = u

⇒ 2x dx = du

⇒ x dx = 1/2 du

∴ I =

As,

∴ Let, u – 3 = A(2u + 2) + B

⇒ u – 3 = 2Au + 2A + B

On comparing both sides –

We have,

2A = 1 ⇒ A = 1/2

2A + B = –3 ⇒ B = –3–2A = –4

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ – 4I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let v = u² + 2u – 4 ⇒ dv = (2u + 2)du

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator.

∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with

∴ I₂ = …eqn 3

From eqn 1:

I = I₁ – 4I₂

Using eqn 2 and eqn 3:

I =

I =

Putting value of u in I:

I =

---

Exercise 19.20

Question 1.

Evaluate the following integrals:

Answer:

Given

Expressing the integral

Consider

By partial fraction decomposition,

⇒ 2x + 1 = Ax + B(x – 1)

⇒ 2x + 1 = Ax + Bx – B

⇒ 2x + 1 = (A + B)x – B

∴ B = -1 and A + B = 2

∴ A = 2 + 1 = 3

Thus,

Consider

Substitute u = x – 1 → dx = du.

We know that

Then,

∴

Then,

We know that

∴

Question 2.

Evaluate the following integrals:

Answer:

Consider

Expressing the integral

Let x² + x – 1 = x² + x – 6 + 5

Consider

Factorizing the denominator,

By partial fraction decomposition,

⇒ 1 = A(x + 3) + B(x – 2)

⇒ 1 = Ax + 3A + Bx – 2B

⇒ 1 = (A + B) x + (3A – 2B)

⇒ Then A + B = 0 … (1)

And 3A – 2B = 1 … (2)

Solving (1) and (2),

2 × (1) → 2A + 2B = 0

1 × (2) → 3A – 2B = 1

5A = 1

∴ A = 1/5

Substituting A value in (1),

⇒ A + B = 0

⇒ 1/5 + B = 0

∴ B = -1/5

Thus,

Let x – 2 = u → dx = du

And x + 3 = v → dx = dv.

We know that

Then,

We know that

∴

Question 3.

Evaluate the following integrals:

Answer:

Given

Rewriting, we get

Expressing the integral

Consider

By partial fraction decomposition,

⇒ x – 2 = A (2x – 1) + Bx

⇒ x – 2 = 2Ax – A + Bx

⇒ x – 2 = (2A + B) x – A

∴ A = 2 and 2A + B = 1

∴ B = 1 – 4 = -3

Thus,

Consider

We know that

And consider

Let u = 2x – 1 → dx = 1/2 du

We know that

Then,

Then,

We know that

Question 4.

Evaluate the following integrals:

Answer:

Consider

Expressing the integral

Consider

Let and split,

Consider

Let

We know that

Now consider

By partial fraction decomposition,

⇒ 1 = A (x – 2) + B (x – 3)

⇒ 1 = Ax – 2A + Bx – 3B

⇒ 1 = (A + B) x – (2A + 3B)

⇒ A + B = 0 and 2A + 3B = -1

Solving the two equations,

⇒ 2A + 2B = 0

2A + 3B = -1

-B = 1

∴ B = -1 and A = 1

Consider

Let u = x – 3 → dx = du

We know that

Similarly

Let u = x – 2 → dx = du

We know that

Then,

Then,

Then,

We know that

Question 5.

Evaluate the following integrals:

Answer:

Given

Expressing the integral

Consider

Let and split,

Consider

Let

We know that

Now consider

By partial fraction decomposition,

⇒ 1 = A (x + 2) + B (x + 5)

⇒ 1 = Ax + 2A + Bx + 5B

⇒ 1 = (A + B) x + (2A + 5B)

⇒ A + B = 0 and 2A + 5B = 1

Solving the two equations,

⇒ 2A + 2B = 0

2A + 5B = 1

-3B = -1

∴ B = 1/3 and A = -1/3

Consider

Let u = x + 2 → dx = du

We know that

Similarly

Let u = x + 5 → dx = du

We know that

Then,

Then,

Then,

We know that

Question 6.

Evaluate the following integrals:

Answer:

Given

Expressing the integral

Consider

Let x = 1/2 (2x – 1) + 1/2 and split,

Consider

Let u = x² – x + 1 → dx = du/2x – 1

We know that

Now consider

Let

We know that

Then,

Now

We know that

Question 7.

Evaluate the following integrals:

Answer:

Given

Expressing the integral

Consider

Let 4x + 1 = 2 (2x + 2) – 3 and split,

Consider

Let

We know that

Now consider

Let u = x + 1 → dx = du

We know that

Then,

Then,

We know that

Question 8.

Evaluate the following integrals:

Answer:

Given

Expressing the integral

Consider

Let and split,

Consider

Let

We know that

Consider

Let

We know that

Then,

Then,

We know that and

Question 9.

Evaluate the following integrals:

Answer:

Given

Expressing the integral

Consider

Let u = 1/2 x → dx = 2du

We know that

Then,

We know that and

Question 10.

Evaluate the following integrals:

Answer:

Given

Expressing the integral

Consider

Let x + 2 = 1/2(2x + 6) – 1 and split,

Consider

Let

We know that

Now consider

Let

We know that

Then,

Then,

We know that

---

Exercise 19.3

Question 1.

Evaluate:

Answer:

Let I = then,

I =

=

=

=

Hence, I = +C

Question 2.

Evaluate:

Answer:

Let I = then,

I=

= +

= +

Hence, I = + +C

Question 3.

Evaluate:

Answer:

Let I = dx

I= dx

We know +C

=

=

Question 4.

Evaluate:

Answer:

Let I =

I =

= dx

=

=

=

=

Hence, I =

Question 5.

Evaluate:

Answer:

Let I =

=

Now Multiply with the conjugate, we get

=

=

= dx

=

=

Hence I= +C

Question 6.

Evaluate:

Answer:

Let I = dx

I = dx

Now, Multiply with the conjugate, we get

=

=

=

=

=

=

=

Hence, I=

Question 7.

Evaluate:

Answer:

Let I =

=

=

=

=

=

Hence, I= +C

Question 8.

Evaluate:

Answer:

Let I = dx

= dx

Now, Multiply with conjugate, we get

= dx

= dx

= dx

=

Hence, I= +C

Question 9.

Evaluate:

Answer:

Let I =

=

=

=

=

Now, Multiply and Divide by 2 we get,

=

=

=

Hence, I= +C

Question 10.

Evaluate:

Answer:

Let I =

⇒

⇒

⇒

⇒

⇒

Hence, I= +C

Question 11.

Evaluate:

Answer:

Let I = dx

= dx

=

=

=

=

Hence, I= +C

Question 12.

Evaluate:

Answer:

Let I =

=

Now, Multiply with the conjugate we get,

= dx

= dx

=

=

=

=

Hence, I= +C

Question 13.

Evaluate:

Answer:

Let I =

=

Now Multiply with Conjugate,

=

=

=

=

=

=

=

Hence, I= +C

Question 14.

Evaluate:

∫ (e^x + 1)² e^x dx

Answer:

Let I = ∫ (e^x + 1)² e^x dx

Let e^x +1= t = e^x dx = dt

I = ∫ (e^x + 1)² e^x dx

= ∫ t² dt

=

Now, substitute the value of t

Hence, I= +C

Question 15.

Evaluate:

Answer:

Let I =

=

=

Hence, I = +C

Question 16.

Evaluate:

Answer:

Let I = dx

= dx

= dx

= dx

= dx

= dx

= dx

=

=

=

=

Hence, I = +C

Question 17.

Evaluate:

Answer:

Let I =

=

Now, Multiply with the conjugate

=

=

=

=

=

Hence, I = +C

Question 18.

∫ tan²(2x – 3)dx

Answer:

Let I =

=

=

Let 2x -3 = t dx = dt/2

=

= tan t –x

Substitute the value of t

Hence, I= +C

Question 19.

Evaluate:

Answer:

Let I =

=

=

=

=

=

=

=

Hence, I = +C

---

Exercise 19.21

Question 1.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ x = λ (2x + 6) + μ

∴ λ = 1/2 and μ = -3

Let x = 1/2(2x + 6) – 3 and split,

Consider

Let

We know that

Consider

Let u = x + 3 → dx = du

We know that

Then,

Question 2.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ 2x + 1 = λ (2x + 2) + μ

∴ λ = 1 and μ = -1

Let 2x + 1 = 2x + 2 – 1 and split,

Consider

Let

We know that

Consider

Let

We know that

Then,

Question 3.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ x + 1 = λ (-2x + 5) + μ

∴ λ = -1/2 and μ = 7/2

Let x + 1 = – 1/2(–2x + 5) + 7/2

Consider

Let

We know that

Consider

Let

We know that

Then,

Question 4.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ 6x – 5 = λ (6x – 5) + μ

∴ λ = 1 and μ = 0

Let

We know that

Question 5.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ 3x + 1 = λ (-2x – 2) + μ

∴ λ = -3/2 and μ = -2

Let 3x + 1 = – (3/2)(–2x – 2) – 2

Consider

Let

We know that

Consider

Let

We know that

Then,

Question 6.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ x = λ (-2x + 1) + μ

∴ λ = -1/2 and μ = -1/2

Let x = -1/2(-2x + 1) – 1/2 and split,

Consider

Let

We know that

Consider

Let

We know that

Then,

Question 7.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ x + 2 = λ (2x + 2) + μ

∴ λ = 1/2 and μ = 1

Let x + 2 = 1/2(2x + 2) + 1 and split,

Consider

Let

We know that

Consider

Let

We know that

Then,

Question 8.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ x + 2 = λ (2x) + μ

∴ λ = 1/2 and μ = 2

Let x + 2 = 1/2(2x) + 2 and split,

Consider

Let

We know that

Consider

We know that

Then,

Question 9.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ x – 1 = λ (2x) + μ

∴ λ = 1/2 and μ = -1

Let x – 1 = 1/2(2x) – 1 and split,

Consider

Let

We know that

Consider

We know that

Then,

Question 10.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ x = λ (2x + 1) + μ

∴ λ = 1/2 and μ = -1/2

Let x = 1/2(2x + 1) – 1/2 and split,

Consider

Let

We know that

Consider

Let

We know that

Then,

Question 11.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ x + 1 = λ (2x) + μ

∴ λ = 1/2 and μ = 1

Let x + 1 = 1/2(2x) + 1 and split,

Consider

Let

We know that

Consider

We know that

Then,

Question 12.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ 2x + 5 = λ (2x + 2) + μ

∴ λ = 1 and μ = 3

Let 2x + 5 = 2x + 2 + 3 and split,

Consider

Let

We know that

Consider

Let

We know that

Then,

Question 13.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ 3x + 1 = λ (-2x – 2) + μ

∴ λ = -3/2 and μ = -2

Let 3x + 1 = – (3/2)(–2x – 2) – 2

Consider

Let

We know that

Consider

Let

We know that

Then,

Question 14.

Evaluate the following integrals:

Answer:

Given

Rationalizing the denominator,

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ -x + 1 = λ (-2x) + μ

∴ λ = 1/2 and μ = 1

Let -x + 1 = 1/2(-2x) + 1 and split,

Consider

Let

We know that

Consider

We know that

Then,

Question 15.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ 2x + 1 = λ (2x + 4) + μ

∴ λ = 1 and μ = -3

Let 2x + 1 = 2x + 4 – 3 and split,

Consider

Let

We know that

Consider

Let u = x + 2 → dx = du

We know that

Then,

Question 16.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ 2x + 3 = λ (2x + 4) + μ

∴ λ = 1/2 and μ = -1

Let 2x + 3 = 2x + 4 – 1 and split,

Consider

Let

We know that

Consider

Let u = x + 2 → dx = du

We know that

Then,

Question 17.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ 5x + 3 = λ (2x + 4) + μ

∴ λ = 5/2 and μ = -7

Let and split,

Consider

Let

We know that

Consider

Let

We know that

Then,

Question 18.

Evaluate the following integrals:

Answer:

Given

Integral is of form

Writing numerator as

⇒ px + q = λ(2ax + b) + μ

⇒ x + 2 = λ (2x + 2) + μ

∴ λ = 1/2 and μ = 1

Let x + 2 = 1/2(2x + 2) + 1 and split,

Consider

Let

We know that

Consider

Let

We know that

Then,

---

Exercise 19.22

Question 1.

Evaluate the following integrals:

Answer:

Given

Dividing the numerator and denominator of the given integrand by cos²x, we get

Putting tanx = t and sec²x dx = dt, we get

We know that

Question 2.

Evaluate the following integrals:

Answer:

Given

Dividing the numerator and denominator of the given integrand by cos²x, we get

Putting tanx = t and sec²x dx = dt, we get

We know that

Question 3.

Evaluate the following integrals:

..

Answer:

Given

We know that sin 2x = 2 sin x cos x

Dividing the numerator and denominator by cos² x,

Replacing sec² x in denominator by 1 + tan² x,

Putting tan x = t so that sec² x dx = dt,

We know that

Question 4.

Evaluate the following integrals:

Answer:

Given

Dividing numerator and denominator by cos²x,

Replacing sec²x by 1 + tan²x in denominator,

Putting tan x = t and sec²x dx = dt, we get

We know that

Question 5.

Evaluate the following integrals:

Answer:

Given

Divide numerator and denominator by cos²x,

Replacing sec² x in denominator by 1 + tan² x,

Putting tan x = t so that sec² x dx = dt,

We know that

Question 6.

Evaluate the following integrals:

Answer:

Given

Divide numerator and denominator by cos²x,

Replacing sec² x in denominator by 1 + tan² x,

Putting tan x = t so that sec² x dx = dt,

We know that

Question 7.

Evaluate the following integrals:

Answer:

Given

Dividing the numerator and denominator by cos²x,

Putting tan x = t so that sec²x dx = dt.

We know that

Question 8.

Evaluate the following integrals:

Answer:

Given

Dividing the numerator and denominator by cos⁴ x,

Putting tan² x = t so that 2tan x sec² x dx = dt

We know that

Question 9.

Evaluate the following integrals:

.w.

Answer:

Given

Dividing the numerator and denominator by cos² x,

Putting tan x + 2 = t so that sec² x dx = dt,

We know that

Question 10.

Evaluate the following integrals:

Answer:

Given

We know that sin 2x = 2 sin x cos x

Dividing numerator and denominator by cos² x,

Putting tan x = t so that sec² x dx = dt,

We know that

Question 11.

Evaluate the following integrals:

Answer:

Given

We know that cos 2x = 1 – 2sin² x.

Dividing numerator and denominator by cos²x,

Replacing sec²x in denominator by 1 + tan²x,

Putting tan x = t so that sec²x dx = dt,

We know that

---

Exercise 19.23

Question 1.

Evaluate the following integrals:

Answer:

Given

We know that

Replacing 1 + tan²x/2 in numerator by sec²x/2,

Putting tanx/2 = t and sec²(x/2)dx = 2dt,

We know that

Question 2.

Evaluate the following integrals:

Answer:

Given

We know that

Replacing 1 + tan²x/2 in numerator by sec²x/2,

Putting tanx/2 = t and sec²(x/2)dx = 2dt,

We know that

Question 3.

Evaluate the following integrals:

Answer:

Given

We know that

Replacing 1 + tan²x/2 in numerator by sec²x/2,

Putting tanx/2 = t and sec²(x/2)dx = 2dt,

We know that

Question 4.

Evaluate the following integrals:

Answer:

Given

We know that

Replacing 1 + tan²x/2 in numerator by sec²x/2,

Putting,

We know that

Question 5.

Evaluate the following integrals:

Answer:

Given

We know that and

Replacing 1 + tan²x/2 in numerator by sec²x/2 and putting tan x/2 = t and sec² x/2 dx = 2dt,

We know that

Question 6.

Evaluate the following integrals:

Answer:

Given

We know that and

Replacing 1 + tan²x/2 in numerator by sec²x/2 and putting tan x/2 = t and sec² x/2 dx = 2dt,

We know that

Question 7.

Evaluate the following integrals:

Answer:

Given

We know that and

Replacing 1 + tan²x/2 in numerator by sec²x/2 and putting tan x/2 = t and sec² x/2 dx = 2dt,

We know that

Question 8.

Evaluate the following integrals:

Answer:

Given

We know that and

Replacing 1 + tan²x/2 in numerator by sec²x/2 and putting tan x/2 = t and sec² x/2 dx = 2dt,

We know that

Question 9.

Evaluate the following integrals:

Answer:

Given

We know that and

Replacing 1 + tan²x/2 in numerator by sec²x/2 and putting tan x/2 = t and sec² x/2 dx = 2dt,

We know that

Question 10.

Evaluate the following integrals:

Answer:

Given

We know that

Replacing 1 + tan²x/2 in numerator by sec²x/2,

Putting tanx/2 = t and sec²(x/2)dx = 2dt,

We know that

Question 11.

Evaluate the following integrals:

Answer:

Given

We know that and

Replacing 1 + tan²x/2 in numerator by sec²x/2 and putting tan x/2 = t and sec² x/2 dx = 2dt,

We know that

Question 12.

Evaluate the following integrals:

Answer:

Given

We know that and

Replacing 1 + tan²x/2 in numerator by sec²x/2 and putting tan x/2 = t and sec² x/2 dx = 2dt,

We know that

Question 13.

Evaluate the following integrals:

Answer:

Given

Let √3 = r cosθ and 1 = r sinθ

And tan θ = 1/√3 → θ = π/6

We know that

Question 14.

Evaluate the following integrals:

Answer:

Given

Let 1 = r cosθ and √3 = r sinθ

And tan θ = √3 → θ = π/3

We know that

Question 15.

Evaluate the following integrals:

Answer:

Given

We know that and

Replacing 1 + tan²x/2 in numerator by sec²x/2 and putting tan x/2 = t and sec² x/2 dx = 2dt,

We know that

---

Exercise 19.24

Question 1.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

C = 0

A – B = 0 ⇒ A = B

B + A = 1 ⇒ 2A = 1 ⇒ A = 1/2

∴ A = B = 1/2

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, u = sin x – cos x ⇒ du = (cos x + sin x)dx

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

∴ I₂ = …..equation 3

From equation 1 ,2 and 3 we have:

I =

∴ I =

Question 2.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

C = 0

B – A = 1 ⇒ A = B - 1

B + A = 0 ⇒ 2B - 1 = 0 ⇒ B = 1/2

∴ A = B - 1 = -1/2

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, u = cos x – sin x ⇒ du = -(cos x + sin x)dx

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

∴ I₂ = …..equation 3

From equation 1 ,2 and 3 we have:

I =

∴ I =

Question 3.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

3B+ C = 3

B + 2A = 2

2B - A = 4

On solving for A ,B and C we have:

A = 0, B = 2 and C = -3

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

So, I₁ reduces to:

I₁ = …..equation 2

As, I₂ =

To solve the integrals of the form

To apply substitution method we take following procedure.

We substitute:

∴ I₂ =

⇒ I₂ =

⇒ I₂ =

⇒ I₂ =

Let, t = ⇒

∴ I₂ =

As, the denominator is polynomial without any square root term. So one of the special integral will be used to solve I₂.

I₂ =

⇒ I₂ =

∴ I₂ = {∵ a² + 2ab + b² = (a+b)²}

As, I₂ matches with the special integral form

I₂ =

Putting value of t we have:

∴ I₂ = + C₂ ……equation 3

From equation 1,2 and 3:

I = + C₂

∴ I = + C ….ans

Question 4.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

C = 0

Bp + Aq = 1

Bq + Ap = 0

On solving above equations, we have:

A = B = and C = 0

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, u = pcos x + qsin x ⇒ du = (-psin x + qcos x)dx

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

∴ I₂ = …..equation 3

From equation 1 ,2 and 3 we have:

I =

∴ I =

Question 5.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

3B+ C = 6

2B + A = 5

B - 2A = 0

On solving for A ,B and C we have:

A = 1, B = 2 and C = 0

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, 2 cos x + sin x + 3 = u

⇒ (-2sin x + cos x)dx = du

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

⇒ I₂ = 2 …..equation 3

From equation 1, 2 and 3 we have:

I = +

∴ I =

Question 6.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

C = 0

3B - 4A = 2

4B + 3A = 3

On solving for A ,B and C we have:

A = 1/25 , B = 18/25 and C = 0

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, 4 cos x + 3sin x = u

⇒ (-4sin x + 3cos x)dx = du

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

⇒ I₂ = …..equation 3

From equation 1, 2 and 3 we have:

I =

∴ I =

Question 7.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

C = 0

3B - 4A = 1

4B + 3A = 0

On solving for A ,B and C we have:

A = -4/25 , B = 3/25 and C = 0

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, 4 cos x + 3sin x = u

⇒ (-4sin x + 3cos x)dx = du

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

⇒ I₂ = …..equation 3

From equation 1, 2 and 3 we have:

I =

∴ I =

Question 8.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

C = 0

3B - 4A = 2

4B + 3A = 3

On solving for A ,B and C we have:

A = 1/25 , B = 18/25 and C = 0

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, 4 cos x + 3sin x = u

⇒ (-4sin x + 3cos x)dx = du

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

⇒ I₂ = …..equation 3

From equation 1, 2 and 3 we have:

I =

∴ I =

Question 9.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

C = 0

3B - 4A = 0

4B + 3A = 1

On solving for A ,B and C we have:

A = 3/25 , B = 4/25 and C = 0

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, 4 cos x + 3sin x = u

⇒ (-4sin x + 3cos x)dx = du

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

⇒ I₂ = …..equation 3

From equation 1, 2 and 3 we have:

I =

∴ I =

Question 10.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

C = 0

2B - 3A = 1

3B + 2A = 8

On solving for A ,B and C we have:

A = 1 , B = 2 and C = 0

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, 3 cos x + 2 sin x = u

⇒ (-3sin x + 2cos x)dx = du

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

⇒ I₂ = …..equation 3

From equation 1, 2 and 3 we have:

I =

∴ I =

Question 11.

Evaluate the integral

Answer:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I =

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form

Then substitute numerator as -

Where A, B and C are constants

We have, I =

As I matches with the form described above, So we will take the steps as described.

∴

⇒ {

⇒

Comparing both sides we have:

C = 0

5B - 4A = 4

4B + 5A = 5

On solving for A ,B and C we have:

A =9/41, B = 40/41 and C = 0

Thus I can be expressed as:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

Let, 4 cos x + 5sin x = u

⇒ (-4sin x + 5cos x)dx = du

So, I₁ reduces to:

I₁ =

∴ I₁ = …..equation 2

As, I₂ =

⇒ I₂ = …..equation 3

From equation 1, 2 and 3 we have:

I =

∴ I =

---

Exercise 19.25

Question 1.

Evaluate the following integrals:

∫ x cos x dx

Answer:

Let

We know that,

Using integration by parts,

We have,

Question 2.

Evaluate the following integrals:

∫ log (x + 1) dx

Answer:

Let

That is,

Using integration by parts,

We know that,

Question 3.

Evaluate the following integrals:

∫ x³ log x dx

Answer:

Let

Using integration by parts,

We have,

Question 4.

Evaluate the following integrals:

∫ xe^x dx

Answer:

Let

Using integration by parts,

We know that ,

Question 5.

Evaluate the following integrals:

∫ xe^2x dx

Answer:

Let

Using integration by parts,

We know that ,

Question 6.

Evaluate the following integrals:

∫ x² e^–x dx

Answer:

Let

Using integration by parts,

We know that,

Using integration by parts in second integral,

Question 7.

Evaluate the following integrals:

∫ x² cos x dx

Answer:

Let

Using integration by parts,

We know that,and

We know that,

Question 8.

Evaluate the following integrals:

∫ x² cos 2x dx

Answer:

Let

Using integration by parts,

We know that,

Then,

Using integration by parts in

Question 9.

Evaluate the following integrals:

∫ x sin 2x dx

Answer:

Let

Using integration by parts,

We know that, and

Question 10.

Evaluate the following integrals:

Answer:

Let

It can be written as,

Using integration by parts,

We know that,and

Question 11.

Evaluate the following integrals:

∫ x² cos x dx

Answer:

Let

Using integration by parts,

We know that,

Using integration by parts in second integral,

Question 12.

Evaluate the following integrals:

∫ x cosec² x dx

Answer:

Let

Using integration by parts,

We know that, and

Question 13.

Evaluate the following integrals:

∫ x cos² x dx

Answer:

Let

Using integration by parts,

We know that,

We know that,

Question 14.

Evaluate the following integrals:

∫ xⁿ log x dx

Answer:

Let

Using integration by parts,

We know that,

and

We know that,

Question 15.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

We know that,

Question 16.

Evaluate the following integrals:

∫ x² sin² x dx

Answer:

Let

We know that,

Using integration by parts,

Using integration by parts in second integral,

Using integration by parts again,

Question 17.

Evaluate the following integrals:

Answer:

Let

Put x²=t

2xdx=dt

Using integration by parts,

We have,

Substitute value for t,

Question 18.

Evaluate the following integrals:

∫ x³ cos x² dx

Answer:

Let

Put x²=t

2xdx=dt

Using integration by parts,

Substitute value for t,

Question 19.

Evaluate the following integrals:

∫ x sin x cos x dx

Answer:

Let

We know that,

Using integration by parts,

We have,

and

Question 20.

Evaluate the following integrals:

∫ sin x log (cos x) dx

Answer:

Let

Put cos x =t

-sinx dx=dt

Using integration by parts,

Replace t by cos x

Question 21.

Evaluate the following integrals:

∫ (log x)² x dx

Answer:

Let

Using integration by parts,

Using integration by integration by parts in second integral,

We know that, and

Question 22.

Evaluate the following integrals:

Answer:

Let

dx=2tdt

Using integration by parts,

Replace the value of t

Question 23.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

We know that, and

Replace the value of t,

Question 24.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

Question 25.

Evaluate the following integrals:

∫ log₁₀ x dx

Answer:

Let

Using integration by parts,

We know that

Question 26.

Evaluate the following integrals:

∫ cos √x dx

Answer:

Let

dx=2tdt

Using integration by parts,

Replace the value of t,

Question 27.

Evaluate the following integrals:

Answer:

Let

Let

Also,

cos t =x

Thus,

Now let us solve this by ‘by parts’ method

Using integration by parts,

Let

U=t; du=dt

Thus,

Substituting

Question 28.

Evaluate the following integrals:

Answer:

We know that integration by parts is given by:

Choosing log x as first function and as second function we get,

+c

+ c

Question 29.

Evaluate the following integrals:

∫ cosec³ x dx

Answer:

Let

Using integration by parts,

We know that,

Using integration by parts,

Question 30.

Evaluate the following integrals:

∫ sec^–1 √x dx

Answer:

Let

dx=2tdt

Using integration by parts,

We know that,

Substitute value for t,

Question 31.

Evaluate the following integrals:

∫ sin^–1 √x dx

Answer:

Let

dx=2tdt

Using integration by parts,

We know that,

t=sin u;dt=cos u du

There fore,

Question 32.

Evaluate the following integrals:

∫ x tan² x dx

Answer:

Let

Using integration by parts,

We know that,

Question 33.

Evaluate the following integrals:

Answer:

Let it can be written n terms of cos x

Using integration by parts,

Question 34.

Evaluate the following integrals:

∫ (x + 1)e^x log(xe^x) dx

Answer:

Let

Using integration by parts,

Substitute value for t,

Question 35.

Evaluate the following integrals:

∫ sin^–1 (3x – 4x³) dx

Answer:

Let ∫ sin^–1 (3x – 4x³) dx

We know that

We know that,

Using integration by parts,

Question 36.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

We know that,

Question 37.

Evaluate the following integrals:

Answer:

Let

We know that,

We know that,

Using integration by parts,

Question 38.

Evaluate the following integrals:

∫ x² sin^–1 x dx

Answer:

Let

Using integration by parts,

Let 1-x²=t²

-2x dx=2t dt

-x dx=t dt

Question 39.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

Where,

-2xdx=2tdt

Question 40.

Evaluate the following integrals:

Answer:

Let

We know that,

Using integration by parts,

Question 41.

Evaluate the following integrals:

∫ cos^–1 (4x³ – 3x) dx

Answer:

Let

We know that,

Using integration by parts,

Question 42.

Evaluate the following integrals:

Answer:

Let

Let x=tan t

dx=sec²t dt

We know that

Using integration by parts,

Question 43.

Evaluate the following integrals:

Answer:

Let

We know that,

Using integration by parts,

Question 44.

Evaluate the following integrals:

∫ (x + 1) log x dx

Answer:

Let

Using integration by parts,

We know that,

Question 45.

Evaluate the following integrals:

∫ x² tan^–1 x dx

Answer:

Let

Using integration by parts,

Taking inverse function as first function and algebraic function as second function,

Question 46.

Evaluate the following integrals:

∫ (e^{log x} + sin x) cos x dx

Answer:

Let

Using integration by parts,

Question 47.

Evaluate the following integrals:

Answer:

Let

We know that,

Using integration by parts,

Substitute value for t

Question 48.

Evaluate the following integrals:

∫ tan^–1 (√x) dx

Answer:

Let

x=t²

dx=2tdt

Using integration by parts,

We know that,

Question 49.

Evaluate the following integrals:

∫ x³ tan^–1 x dx

Answer:

Let

Using integration by parts,

We know that,

Question 50.

Evaluate the following integrals:

∫ x sin x cos 2x dx

Answer:

Let

Using integration by parts,

Using integration by parts,

Question 51.

Evaluate the following integrals:

∫ (tan^–1 x²) x dx

Answer:

Let

X²=t

2xdx=dt

Using integration by parts,

We know that,

Question 52.

Evaluate the following integrals:

Answer:

Let

We are splitting this in to two functions

First we find the integral of:

Put 1-x²=t

-2xdx=dt

Using integration by parts,

Question 53.

Evaluate the following integrals:

∫ sin³ √x dx

Answer:

Let

dx=2tdt

Using integration by parts,

Question 54.

Evaluate the following integrals:

∫ x sin³ x dx

Answer:

Let

We know that,

Using integration by parts,

Question 55.

Evaluate the following integrals:

∫ cos³ √x dx

Answer:

Let

dx=2tdt

let

we know that,

Using integration by parts,

Question 56.

Evaluate the following integrals:

∫ x cos³ x dx

Answer:

Let

we know that,

Using integration by parts,

Question 57.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

Question 58.

Evaluate the following integrals:

Answer:

Let

Let

Using integration by parts,

Question 59.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

Question 60.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

We know that,

---

Exercise 19.26

Question 1.

Evaluate the following integrals:

∫ e^x (cos x – sin x) dx

Answer:

Let

Using integration by parts,

We know that,

Question 2.

Evaluate the following integrals:

Answer:

Let

Integrating by parts

We know that,

Question 3.

Evaluate the following integrals:

Answer:

Let

We know that,

Let

We know that,

From equation(1), we obtain

Question 4.

Evaluate the following integrals:

∫ e^x (cot x – cosec² x) dx

Answer:

Let

Integrating by parts,

Question 5.

Evaluate the following integrals:

Answer:

Let

Integrating by parts,

Question 6.

Evaluate the following integrals:

∫ e^x sec x (1 + tan x) dx

Answer:

Let

Integrating by parts,

Question 7.

Evaluate the following integrals:

∫ e^x (tan x – log cos x) dx

Answer:

Let

Integrating by parts,

Question 8.

Evaluate the following integrals:

∫ e^x [sec x + log (sec x + tan x)] dx

Answer:

Let

Integrating by parts

Question 9.

Evaluate the following integrals:

∫ e^x (cot x + log sin x) dx

Answer:

Let

Integrating by parts

Question 10.

Evaluate the following integrals:

Answer:

Let

Integrating by parts

Question 11.

Evaluate the following integrals:

Answer:

Let

Integrating by parts,

Question 12.

Evaluate the following integrals:

Answer:

Let

We know that,

Question 13.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

Question 14.

Evaluate the following integrals:

Answer:

Let

Integrating by parts

Question 15.

Evaluate the following integrals:

Answer:

Let

We know that

Here,

Question 16.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

Question 17.

Evaluate the following integrals:

Answer:

Let

Using integration by parts,

Question 18.

Evaluate the following integrals:

Answer:

Let

Integrating by parts

Question 19.

Evaluate the following integrals:

∫ e^2x (– sin x + 2 cos x) dx

Answer:

Let

Applying by parts in the second integral,

Question 20.

Evaluate the following integrals:

Answer:

Let

and we know that,

Question 21.

Evaluate the following integrals:

Answer:

Let

We know that,

let

Question 22.

Evaluate the following integrals:

∫ {tan (log x) + sec² (log x)} dx

Answer:

Let

We know that,

Question 23.

Evaluate the following integrals:

Answer:

Let

Let

We know that,

Question 24.

Evaluate the following integrals:

Answer:

Let

We have,

Using integration by parts,

That is,

I=I₁+I₂

Consider

Let

Thus,

---

Exercise 19.27

Question 1.

Evaluate the following integrals:

∫ e^ax cos bx dx

Answer:

Let

Integrating by parts,

Question 2.

Evaluate the following integrals:

∫ e^ax sin (bx + c) dx

Answer:

Let

Question 3.

Evaluate the following integrals:

∫ cos (log x) dx

Answer:

Let

Let log x=t

dx=xdt

We know that,

Hence, a=1, b=1

So ,

Hence,

Question 4.

Evaluate the following integrals:

∫ e^2x cos (3x + 4) dx

Answer:

Let

Integrating by parts

Hence,

Question 5.

Evaluate the following integrals:

∫ e^2x sin x cos x dx

Answer:

Let

We know that,

Question 6.

Evaluate the following integrals:

e^2x sin x dx

Answer:

Let

Integrating by parts,

Again integrating by parts,

Question 7.

Evaluate the following integrals:

∫ e^2x sin (3x + 1) dx

Answer:

Let I = ∫ e^2x sin (3x + 1) dx

Now Integrating by parts choosing sin (3x + 1) as first function and e^2x as second function we get,

Now again integrating by parts by taking cos(3x + 1) as first function and e^2x as second function we get,

Therefore,

Question 8.

Evaluate the following integrals:

∫ e^x sin² x dx

Answer:

Let

Using integration by parts,

We know that,

Question 9.

Evaluate the following integrals:

Answer:

Let

We know that

Question 10.

Evaluate the following integrals:

∫ e^2x cos² x dx

Answer:

Let

We know that,

Question 11.

Evaluate the following integrals:

∫ e^–2x sin x dx

Answer:

Let

We know that,

Question 12.

Evaluate the following integrals:

Answer:

Let

We know that,

---

Exercise 19.28

Question 1.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

∴ I =

Using a² – 2ab + b² = (a - b)²

We have:

I =

As I match with the form:

∴ I =

⇒ I =

Question 2.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of the method of substitution along with a method of integration by parts. By the method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

∴ I =

Using a² – 2ab + b² = (a - b)²

We have:

I =

As I match with the form:

∴ I =

⇒ I =

Question 3.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

∴ I =

Using a² – 2ab + b² = (a - b)²

We have:

I =

As I match with the form:

∴ I =

⇒ I =

⇒ I =

Question 4.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of the method of substitution along with a method of integration by parts. By the method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

Let, sin x = t

Differentiating both sides:

⇒ cos x dx = dt

Substituting sin x with t, we have:

∴ I =

As I match with the form:

∴ I =

Putting the value of t i.e. t = sin x

⇒ I =

Question 5.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

Let, e^x = t

Differentiating both sides:

⇒ e^x dx = dt

Substituting e^x with t, we have:

We have:

I =

As I match with the form:

∴ I =

⇒ I =

Putting the value of t back:

⇒ I =

Question 6.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of the method of substitution along with a method of integration by parts. By the method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

∴ I =

As I match with the form:

∴ I =

Question 7.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of the method of substitution along with a method of integration by parts. By the method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

We have:

I =

⇒ I =

As I match with the form:

∴ I =

⇒ I =

Question 8.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

We have:

I =

⇒ I =

As I match with the form:

∴ I =

⇒ I =

Question 9.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

∴ I =

Using a² + 2ab + b² = (a + b)²

We have:

I =

As I match with the form:

∴ I =

⇒ I =

⇒ I =

Question 10.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

∴ I =

Using a² + 2ab + b² = (a + b)²

We have:

I =

As I match with the form:

∴ I =

⇒ I =

⇒ I =

Question 11.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I = =

Let, x² = t

Differentiating both sides:

⇒ 2x dx = dt ⇒ x dx = 1/2 dt

Substituting x² with t, we have:

We have:

I =

As I match with the form:

∴ I =

⇒ I =

Putting the value of t back:

⇒ I =

⇒ I =

Question 12.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

Let, x³ = t

Differentiating both sides:

⇒ 3x² dx = dt

⇒ x² dx = 1/3 dt

Substituting x³ with t, we have:

∴ I =

As I match with the form:

∴ I =

Putting the value of t i.e. t = x³

⇒ I =

Question 13.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

Let, log x = t

Differentiating both sides:

⇒

Substituting (log x) with t, we have:

We have:

I =

As I match with the form:

∴ I =

Putting the value of t back:

⇒ I =

Question 14.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

∴ I =

Using a² – 2ab + b² = (a - b)²

We have:

I =

As I match with the form:

∴ I =

⇒ I =

Question 15.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

∴ I =

As I match with the form:

∴ I =

Question 16.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

We have:

I =

Using a² – 2ab + b² = (a-b)²

I =

As I match with the form:

∴ I =

⇒ I =

Question 17.

Evaluate the integral:

Answer:

Key points to solve the problem:

• Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have

• To solve the integrals of the form: after applying substitution and integration by parts we have direct formulae as described below:

Let, I =

∴ I =

Using a² – 2ab + b² = (a - b)²

We have:

I =

As I match with the form:

∴ I =

⇒ I =

---

Exercise 19.29

Question 1.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ x + 1 = λ(2x^2-1 – 1 + 0) + μ

⇒ x + 1 = λ(2x – 1) + μ

⇒ x + 1 = 2λx + μ – λ

Comparing the coefficient of x on both sides, we get

2λ = 1 ⇒

Comparing the constant on both sides, we get

μ – λ = 1

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put x² – x + 1 = t

⇒ (2x – 1)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 2.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ x + 1 = λ(2 × 2x^2-1 + 0) + μ

⇒ x + 1 = λ(4x) + μ

⇒ x + 1 = 4λx + μ

Comparing the coefficient of x on both sides, we get

4λ = 1 ⇒

Comparing the constant on both sides, we get

μ = 1

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put 2x² + 3 = t

⇒ (4x)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 3.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ 2x – 5 = λ(0 + 3 – 2x^2-1) + μ

⇒ 2x – 5 = λ(3 – 2x) + μ

⇒ 2x – 5 = –2λx + 3λ + μ

Comparing the coefficient of x on both sides, we get

–2λ = 2 ⇒ λ = –1

Comparing the constant on both sides, we get

3λ + μ = –5

⇒ 3(–1) + μ = –5

⇒ –3 + μ = –5

∴ μ = –2

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put 2 + 3x – x² = t

⇒ (3 – 2x)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 4.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ x + 2 = λ(2x^2-1 + 1 + 0) + μ

⇒ x + 2 = λ(2x + 1) + μ

⇒ x + 2 = 2λx + λ + μ

Comparing the coefficient of x on both sides, we get

2λ = 1 ⇒

Comparing the constant on both sides, we get

λ + μ = 2

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put x² + x + 1 = t

⇒ (2x + 1)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 5.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ 4x + 1 = λ(2x^2-1 – 1 – 0) + μ

⇒ 4x + 1 = λ(2x – 1) + μ

⇒ 4x + 1 = 2λx + μ – λ

Comparing the coefficient of x on both sides, we get

2λ = 4 ⇒

Comparing the constant on both sides, we get

μ – λ = 1

⇒ μ – 2 = 1

∴ μ = 3

Hence, we have 4x + 1 = 2(2x – 1) + 3

Substituting this value in I, we can write the integral as

Let

Now, put x² – x – 2 = t

⇒ (2x – 1)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 6.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ x – 2 = λ(2 × 2x^2-1 – 6 – 0) + μ

⇒ x – 2 = λ(4x – 6) + μ

⇒ x – 2 = 4λx + μ – 6λ

Comparing the coefficient of x on both sides, we get

4λ = 1

Comparing the constant on both sides, we get

μ – 6λ = –2

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put 2x² – 6x + 5 = t

⇒ (4x – 6)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 7.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ x + 1 = λ(2x^2-1 + 1 + 0) + μ

⇒ x + 1 = λ(2x + 1) + μ

⇒ x + 1 = 2λx + λ + μ

Comparing the coefficient of x on both sides, we get

2λ = 1 ⇒

Comparing the constant on both sides, we get

λ + μ = 1

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put x² + x + 1 = t

⇒ (2x + 1)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 8.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ 2x + 3 = λ(2x^2-1 + 4 + 0) + μ

⇒ 2x + 3 = λ(2x + 4) + μ

⇒ 2x + 3 = 2λx + 4λ + μ

Comparing the coefficient of x on both sides, we get

2λ = 2 ⇒ λ = 1

Comparing the constant on both sides, we get

4λ + μ = 3

⇒ 4(1) + μ = 3

⇒ 4 + μ = 3

∴ μ = –1

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put x² + 4x + 3 = t

⇒ (2x + 4)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write x² + 4x + 3 = x² + 2(x)(2) + 2² – 2² + 3

⇒ x² + 4x + 3 = (x + 2)² – 4 + 3

⇒ x² + 4x + 3 = (x + 2)² – 1

⇒ x² + 4x + 3 = (x + 2)² – 1²

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 9.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ 2x – 5 = λ(2x^2-1 – 4 + 0) + μ

⇒ 2x – 5 = λ(2x – 4) + μ

⇒ 2x – 5 = 2λx + μ – 4λ

Comparing the coefficient of x on both sides, we get

2λ = 2 ⇒ λ = 1

Comparing the constant on both sides, we get

μ – 4λ = –5

⇒ μ – 4(1) = –5

⇒ μ – 4 = –5

∴ μ = –1

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put x² – 4x + 3 = t

⇒ (2x – 4)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write x² – 4x + 3 = x² – 2(x)(2) + 2² – 2² + 3

⇒ x² – 4x + 3 = (x – 2)² – 4 + 3

⇒ x² – 4x + 3 = (x – 2)² – 1

⇒ x² – 4x + 3 = (x – 2)² – 1²

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 10.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know

⇒ x = λ(2x^2-1 + 1) + μ

⇒ x = λ(2x + 1) + μ

⇒ x = 2λx + λ + μ

Comparing the coefficient of x on both sides, we get

2λ = 1 ⇒

Comparing the constant on both sides, we get

λ + μ = 0

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put x² + x = t

⇒ (2x + 1)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 11.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ x – 3 = λ(2x^2-1 + 3 + 0) + μ

⇒ x – 3 = λ(2x + 3) + μ

⇒ x – 3 = 2λx + 3λ + μ

Comparing the coefficient of x on both sides, we get

2λ = 1

Comparing the constant on both sides, we get

3λ + μ = –3

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put x² + 3x – 18 = t

⇒ (2x + 3)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 12.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ x + 3 = λ(0 – 4 – 2x^2-1) + μ

⇒ x + 3 = λ(–4 – 2x) + μ

⇒ x + 3 = –2λx + μ – 4λ

Comparing the coefficient of x on both sides, we get

–2λ = 1

Comparing the constant on both sides, we get

μ – 4λ = 3

⇒ μ + 2 = 3

∴ μ = 1

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put 3 – 4x – x² = t

⇒ (–4 – 2x)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write 3 – 4x – x² = –(x² + 4x – 3)

⇒ 3 – 4x – x² = –[x² + 2(x)(2) + 2² – 2² – 3]

⇒ 3 – 4x – x² = –[(x + 2)² – 4 – 3]

⇒ 3 – 4x – x² = –[(x + 2)² – 7]

⇒ 3 – 4x – x² = 7 – (x + 2)²

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 13.

Evaluate the following integrals –

Answer:

Let

Let us assume

We know and derivative of a constant is 0.

⇒ 3x + 1 = λ(0 – 3 – 2×2x^2-1) + μ

⇒ 3x + 1 = λ(–3 – 4x) + μ

⇒ 3x + 1 = –4λx + μ – 3λ

Comparing the coefficient of x on both sides, we get

–4λ = 3

Comparing the constant on both sides, we get

μ – 3λ = 1

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put 4 – 3x – 2x² = t

⇒ (–3 – 4x)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write 4 – 3x – 2x² = –(2x² + 3x – 4)

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

Question 14.

Evaluate the following integrals –

Answer:

Let

Let us assume,

We know and derivative of a constant is 0.

⇒ 2x + 5 = λ(0 – 4 – 3×2x^2-1) + μ

⇒ 2x + 5 = λ(–4 – 6x) + μ

⇒ 2x + 5 = –6λx + μ – 4λ

Comparing the coefficient of x on both sides, we get

–6λ = 2

Comparing the constant on both sides, we get

μ – 4λ = 5

Hence, we have

Substituting this value in I, we can write the integral as

Let

Now, put 10 – 4x – 3x² = t

⇒ (–4 – 6x)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write 10 – 4x – 3x² = –(3x² + 4x – 10)

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,

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Exercise 19.30

Question 1.

Evaluate the following integral:

Answer:

Here the denominator is already factored.

So let

⇒ 2x + 1 = A(x – 2) + B(x + 1)……(ii)

We need to solve for A and B. One way to do this is to pick values for x which will cancel each variable.

Put x = 2 in the above equation, we get

⇒ 2(2) + 1 = A(2 – 2) + B(2 + 1)

⇒ 3B = 5

Now put x = – 1 in equation (ii), we get

⇒ 2( – 1) + 1 = A(( – 1) – 2) + B(( – 1) + 1)

⇒ – 3A = – 1

We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute u = x + 1 ⇒ du = dx and z = x – 2 ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 2.

Evaluate the following integral:

Answer:

Here the denominator is already factored.

So let

⇒ 1 = A(x – 2)(x – 4) + Bx(x – 4) + Cx(x – 2)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = 0 in the above equation, we get

⇒ 1 = A(0 – 2)(0 – 4) + B(0)(0 – 4) + C(0)(0 – 2)

⇒ 1 = 8A + 0 + 0

Now put x = 2 in equation (ii), we get

⇒ 1 = A(2 – 2)(2 – 4) + B(2)(2 – 4) + C(2)(2 – 2)

⇒ 1 = 0 – 4B + 0

Now put x = 4 in equation (ii), we get

⇒ 1 = A(4 – 2)(4 – 4) + B(4)(4 – 4) + C(4)(4 – 2)

⇒ 1 = 0 + 0 + 8C

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute u = x – 4 ⇒ du = dx and z = x – 2 ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

We will take common, we get

Applying the logarithm rule we can rewrite the above equation as

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 3.

Evaluate the following integral:

Answer:

First we simplify numerator, we get

Now we will factorize denominator by splitting the middle term, we get

Now the denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 5 = A(x – 2) + B(x + 3)……(ii)

We need to solve for A and B. One way to do this is to pick values for x which will cancel each variable.

Put x = 2 in the above equation, we get

⇒ 5 = A(2 – 2) + B(2 + 3)

⇒ 5 = 0 + 5B

⇒ B = 1

Now put x = – 3 in equation (ii), we get

⇒ 5 = A(( – 3) – 2) + B(( – 3) + 3)

⇒ 5 = – 5A

⇒ A = – 1

We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute u = x + 3 ⇒ du = dx and z = x – 2 ⇒ dz = dx, so the above equation becomes,

On integrating we get

⇒ x – log|u| + log|z| + C

Substituting back, we get

⇒ x – log|x + 3| + log|x – 2| + C

Applying the logarithm rule, we can rewrite the above equation as

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 4.

Evaluate the following integral:

Answer:

First we simplify numerator, we get

Now the denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 5x + 1 = A(x – 1) + B(x + 2)……(ii)

We need to solve for A and B. One way to do this is to pick values for x which will cancel each variable.

Put x = 1 in the above equation, we get

⇒ 5(1) + 1 = A(1 – 1) + B(1 + 2)

⇒ 6 = 0 + 3B

⇒ B = 2

Now put x = – 2 in equation (ii), we get

⇒ 5( – 2) + 1 = A(( – 2) – 1) + B(( – 2) + 2)

⇒ – 9 = – 3A + 0

⇒ A = 3

We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute u = x + 2 ⇒ du = dx and z = x – 1 ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 5.

Evaluate the following integral:

Answer:

First we simplify numerator, we get

Now the denominator is factorized, so let separate the fraction through partial fraction, hence let

We need to solve for A and B. One way to do this is to pick values for x which will cancel each variable.

Put x = 1 in the above equation, we get

⇒ 2 = A(1 – 1) + B(1 + 1)

⇒ 2 = 0 + 2B

⇒ B = 1

Now put x = – 1 in equation (ii), we get

⇒ 2 = A(( – 1) – 1) + B(( – 1) + 1)

⇒ 2 = – 2A + 0

⇒ A = – 1

We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute u = x + 1 ⇒ du = dx and z = x – 1 ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Applying the logarithm rule we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 6.

Evaluate the following integral:

Answer:

Denominator is already factorized, so let

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = 1 in the above equation, we get

= A(1 – 2)(1 – 3) + B(1 – 1)(1 – 3) + C(1 – 1)(1 – 2)

⇒ 1 = 2A + 0 + 0

Now put x = 2 in equation (ii), we get

= A(2 – 2)(2 – 3) + B(2 – 1)(2 – 3) + C(2 – 1)(2 – 2)

⇒ 4 = 0 – B + 0

⇒ B = – 4

Now put x = 3 in equation (ii), we get

= A(3 – 2)(3 – 3) + B(3 – 1)(3 – 3) + C(3 – 1)(3 – 2)

⇒ 9 = 0 + 0 + 2C

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute u = x – 1 ⇒ du = dx, y = x – 2 ⇒ dy = dx and z = x – 3 ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 7.

Evaluate the following integral:

Answer:

The denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 5x = A(x – 2)(x + 2) + B(x + 1)(x + 2) + C(x + 1)(x – 2)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = – 1 in the above equation, we get

⇒ 5( – 1) = A(( – 1) – 2)(( – 1) + 2) + B(( – 1) + 1)(( – 1) + 2) + C(( – 1) + 1)(( – 1) – 2)

⇒ – 5 = – 3A + 0 + 0

Now put x = – 2 in equation (ii), we get

⇒ 5( – 2) = A(( – 2) – 2)(( – 2) + 2) + B(( – 2) + 1)(( – 2) + 2) + C(( – 2) + 1)(( – 2) – 2)

⇒ – 10 = 0 + 0 + 4C

Now put x = 2 in equation (ii), we get

⇒ 5(2) = A((2) – 2)((2) + 2) + B((2) + 1)((2) + 2) + C((2) + 1)((2) – 2)

⇒ 10 = 0 + 12B + 0

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute u = x + 1 ⇒ du = dx, y = x – 2 ⇒ dy = dx and z = x + 2 ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 8.

Evaluate the following integral:

Answer:

The denominator is factorized, so let separate the fraction through partial fraction, hence let

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = 0 in the above equation, we get

⇒ 0² + 1 = A(0 – 1)(0 + 1) + B(0)(0 + 1) + C(0)(0 – 1)

⇒ 1 = – A + 0 + 0

⇒ A = – 1

Now put x = – 1 in equation (ii), we get

⇒ ( – 1)² + 1 = A(( – 1) – 1)(( – 1) + 1) + B( – 1)(( – 1) + 1) + C( – 1)(( – 1) – 1)

⇒ 2 = 0 + 0 + C

⇒ C = 1

Now put x = 1 in equation (ii), we get

⇒ 1² + 1 = A(1 – 1)(1 + 1) + B(1)(1 + 1) + C(1)(1 – 1)

⇒ 2 = 0 + 2B + 0

⇒ B = 1

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute u = x + 1 ⇒ du = dx, y = x – 1 ⇒ dy = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Applying the rules of logarithm we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 9.

Evaluate the following integral:

Answer:

The denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 2x – 3 = A(x + 1)(2x + 3) + B(x – 1)(2x + 3) + C(x – 1)(x + 1)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = – 1 in the above equation, we get

⇒ 2( – 1) – 3 = A(( – 1) + 1)(2( – 1) + 3) + B(( – 1) – 1)(2( – 1) + 3) + C(( – 1) – 1)(( – 1) + 1)

⇒ – 5 = 0 – 2B + 0

Now put x = 1 in equation (ii), we get

⇒ 2(1) – 3 = A((1) + 1)(2(1) + 3) + B((1) – 1)(2(1) + 3) + C((1) – 1)((1) + 1)

⇒ – 1 = 10A + 0 + 0

Now put in equation (ii), we get

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x + 1 ⇒ du = dx,

y = x – 1 ⇒ dy = dx and

so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 10.

Evaluate the following integral:

Answer:

First we simplify numerator, we will rewrite denominator as shown below

Add and subtract numerator with ( – 6x² + 11x – 6), we get

The denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 6x² – 11x + 6 = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = 1 in the above equation, we get

⇒ 6(1)² – 11(1) + 6 = A(1 – 2)(1 – 3) + B(1 – 1)(1 – 3) + C(1 – 1)(1 – 2)

⇒ 1 = 2A + 0 + 0

Now put x = 2 in equation (ii), we get

6(2)² – 11(2) + 6 = A(2 – 2)(2 – 3) + B(2 – 1)(2 – 3) + C(2 – 1)(2 – 2)

⇒ 8 = 0 – B + 0

⇒ B = – 8

Now put x = 3 in equation (ii), we get

⇒ 6(3)² – 11(3) + 6 = A(3 – 2)(3 – 3) + B(3 – 1)(3 – 3) + C(3 – 1)(3 – 2)

⇒ 27 = 0 + 0 + 2C

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x – 1 ⇒ du = dx,

y = x – 2 ⇒ dy = dx and

z = x – 3 ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 11.

Evaluate the following integral:

Answer:

The denominator is factorized, so let separate the fraction through partial fraction, hence let

We need to solve for A and B.

We will equate similar terms, we get.

2A + B = 0 ⇒ B = – 2A

And A + B = 2 cos x

Substituting the value of B, we get

A – 2A = 2 cos x ⇒ A = – 2 cos x

Hence B = – 2A = – 2( – 2 cos x)

⇒ B = 4cos x

We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

so the above equation becomes,

Now substitute

v = 1 + u ⇒ dv = du

z = 2 + u ⇒ dz = du

So above equation becomes,

On integrating we get

Substituting back, we get

Applying logarithm rule, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 12.

Evaluate the following integral:

Answer:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

By equating similar terms, we get

A + C = 0 ⇒ A = – C ………..(iii)

B + D = 0 ⇒ B = – D …………(iv)

3A + C = 2

⇒ 3( – C) + C = 2 (from equation(iii))

⇒ C = – 1

So equation(iii) becomes A = 1

And also 3B + D = 0 (from equation (ii))

⇒ 3( – D) + D = 0 (from equation (iv))

⇒ D = 0

So equation (iv) becomes, B = 0

We put the values of A, B, C and D values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

so the above equation becomes,

On integrating we get

Substituting back, we get

Applying the logarithm rule we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 13.

Evaluate the following integral:

Answer:

Let substitute, so the given equation becomes

Denominator is factorised, so let separate the fraction through partial fraction, hence let

⇒ 1 = A(2 + u) + Bu……(ii)

We need to solve for A and B. One way to do this is to pick values for x which will cancel each variable.

Put u = – 2 in above equation, we get

⇒ 1 = A(2 + ( – 2)) + B( – 2)

⇒ 1 = – 2B

Now put u = 0 in equation (ii), we get

⇒ 1 = A(2 + 0) + B(0)

⇒ 1 = 2A + 0

We put the values of A and B values back into our partial fractions in equation (ii) and replace this as the integrand. We get

Split up the integral,

Let substitute

z = 2 + u ⇒ dz = du, so the above equation becomes,

On integrating we get

Substituting back the value of z, we get

Now substitute back the value of u, we get

Applying the rules of logarithm we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 14.

Evaluate the following integral:

Answer:

Denominator is factorised, so let separate the fraction through partial fraction, hence let

We need to solve for A, B, C and D. We will equate the like terms we get,

C = 0………….(iii)

A + D = 1⇒ A = 1 – D………(iv)

2A + B + C = 1

⇒ 2(1 – D) + B + 0 = 1 (from equation (iii) and (iv))

⇒ B = 2D – 1……….(v)

2B + D = 1

⇒ 2(2D – 1) + D = 1 (from equation (v), we get

⇒ 4D – 2 + D = 1

⇒ 5D = 3

……………(vi)

Equation (vi) in (v) and (iv), we get

We put the values of A, B, C, and D values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x² + 1 ⇒ du = 2xdx,

y = x + 2 ⇒ dy = dx, so the above equation becomes,

On integrating we get

(the standard integral of )

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 15.

Evaluate the following integral:

, where a, b, c are distinct.

Answer:

Denominator is factorised, so let separate the fraction through partial fraction, hence let

⇒ ax² + bx + c = A(x – b)(x – c) + B(x – a)(x – c) + C(x – a)(x – b)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = a in the above equation, we get

⇒ a(a)² + b(a) + c = A(a – b)(a – c) + B(a – a)(a – c) + C(a – a)(a – b)

⇒ a³ + ab + c = (a – b)(a – c)A + 0 + 0

Now put x = b in equation (ii), we get

⇒ a(b)² + b(b) + c = A(b – b)(b – c) + B(b – a)(b – c) + C(b – a)(b – b)

⇒ ab² + b² + c = 0 + (b – a)(b – c)B + 0

Now put x = c in equation (ii), we get

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x – a ⇒ du = dx,

y = x – b ⇒ dy = dx and

z = x – c ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 16.

Evaluate the following integral:

Answer:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ x = (Ax + B)(x – 1) + (Cx + D)(x² + 1)

⇒ x = Ax² – Ax + Bx – B + Cx² + Cx + Dx² + D

⇒ x = (C) x² + (A + D) x² + (B – A + C)x + (D – B)……(ii)

By equating similar terms, we get

C = 0 ………..(iii)

A + D = 0⇒ A = – D …………(iv)

B – A + C = 1

⇒ B – ( – D) + 0 = 2 (from equation(iii) and (iv))

⇒ B = 2 – D………..(v)

D – B = 0 ⇒ D – (2 – D) = 0 ⇒ 2D = 2 ⇒ D = 1

So equation(iv) becomes A = – 1

So equation (v) becomes, B = 2 – 1 = 1

We put the values of A, B, C, and D values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

so the above equation becomes,

On integrating we get

(the standard integral of )

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 17.

Evaluate the following integral:

Answer:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 1 = A(x + 1)(x + 2) + B(x – 1)(x + 2) + C(x – 1)(x + 1)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = 1 in the above equation, we get

⇒ 1 = A(1 + 1)(1 + 2) + B(1 – 1)(1 + 2) + C(1 – 1)(1 + 1)

⇒ 1 = 6A + 0 + 0

Now put x = – 1 in equation (ii), we get

⇒ 1 = A( – 1 + 1)( – 1 + 2) + B( – 1 – 1)( – 1 + 2) + C( – 1 – 1)( – 1 + 1)

⇒ 1 = 0 – 2B + 0

Now put x = – 2 in equation (ii), we get

⇒ 1 = A( – 2 + 1)( – 2 + 2) + B( – 2 – 1)( – 2 + 2) + C( – 2 – 1)( – 2 + 1)

⇒ 1 = 0 + 0 + 3C

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x – 1 ⇒ du = dx,

y = x + 1 ⇒ dy = dx and

z = x + 2 ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 18.

Evaluate the following integral:

Answer:

Denominator is factorised, so let separate the fraction through partial fraction, hence let

By equating similar terms, we get

A + C = 0 ⇒ A = – C ………..(iii)

B + D = 1⇒ B = 1 – D…………(iv)

9A + 4C = 0

⇒ 9( – C) + 4C = 0 (from equation(iii))

⇒ C = 0………..(v)

So equation(iv) becomes

So equation (iii) becomes, A = 0

We put the values of A, B, C, and D values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

in first partthe

in second parthe t

so the above equation becomes,

On integrating we get

(the standard integral of )

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 19.

Evaluate the following integral:

Answer:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 5x² – 1 = A(x – 1)(x + 1) + Bx(x + 1) + Cx(x – 1)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = 0 in the above equation, we get

⇒ 5(0)² – 1 = A(0 – 1)(0 + 1) + B(0)(0 + 1) + C(0)(0 – 1)

⇒ A = 1

Now put x = 1 in equation (ii), we get

⇒ 5(1)² – 1 = A(1 – 1)(1 + 1) + B(1)(1 + 1) + C(1)(1 – 1)

⇒ 4 = 0 + 2B + 0

⇒ B = 2

Now put x = – 1 in equation (ii), we get

⇒ 5( – 1)² – 1 = A( – 1 – 1)( – 1 + 1) + B( – 1)( – 1 + 1) + C( – 1)( – 1 – 1)

⇒ 4 = 0 + 0 + 2C

⇒ C = 2

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x – 1 ⇒ du = dx,

y = x + 1 ⇒ dy = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Applying logarithm rule, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 20.

Evaluate the following integral:

Answer:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ x² + 6x – 8 = A(x – 2)(x + 2) + Bx(x + 2) + Cx(x – 2)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = 0 in the above equation, we get

⇒ 0² + 6(0) – 8 = A(0 – 2)(0 + 2) + B(0)(0 + 2) + C(0)(0 – 2)

⇒ – 8 = – 4A + 0 + 0

⇒ A = 2

Now put x = 2 in equation (ii), we get

⇒ 2² + 6(2) – 8 = A(2 – 2)(2 + 2) + B(2)(2 + 2) + C(2)(2 – 2)

⇒ 8 = 0 + 8B + 0

⇒ B = 1

Now put x = – 2 in equation (ii), we get

⇒ ( – 2)² + 6( – 2) – 8 = A(( – 2) – 2)(( – 2) + 2) + B( – 2)(( – 2) + 2) + C( – 2)(( – 2) – 2)

⇒ – 16 = 0 + 0 + 8C

⇒ C = – 2

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x – 2 ⇒ du = dx,

y = x + 2 ⇒ dy = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Applying logarithm rule, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 21.

Evaluate the following integral:

Answer:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ x² + 1 = A(x – 1)(x + 1) + B(2x + 1)(x + 1) + C(2x + 1)(x – 1)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = 1 in the above equation, we get

⇒ 1² + 1 = A(1 – 1)(1 + 1) + B(2(1) + 1)(1 + 1) + C(2(1) + 1)(1 – 1)

⇒ 2 = 0 + 6B + 0

Now put in equation (ii), we get

Now put x = – 1 in equation (ii), we get

⇒ ( – 1)² + 1 = A( – 1 – 1)( – 1 + 1) + B(2( – 1) + 1)( – 1 + 1) + C(2( – 1) + 1)( – 1 – 1)

⇒ 2 = 0 + 0 + 2C

⇒ C = 1

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x – 1⇒ du = dx,

y = x + 1 ⇒ dy = dx and

z = 2x + 1 ⇒ dz = 2dx so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 22.

Evaluate the following integral:

Answer:

Let substitute, so the given equation becomes

Factorizing the denominator, we get

The denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 1 = A(3u + 2) + B(2u + 1)……(ii)

We need to solve for A and B. One way to do this is to pick values for x which will cancel each variable.

Put in the above equation, we get

Now put in equation (ii), we get

We put the values of A and B values back into our partial fractions in equation (ii) and replace this as the integrand. We get

Split up the integral,

Let substitute

z = 2u + 1 ⇒ dz = 2du and y = 3u + 2⇒ dy = 3du so the above equation becomes,

On integrating we get

Substituting back the value of z, we get

Now substitute back the value of u, we get

Applying the rules of logarithm we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 23.

Evaluate the following integral:

Answer:

Multiply numerator and denominator by x^{n – 1}, we get

Let xⁿ = t ⇒ nx^{n– 1}dx = dt

So the above equation becomes,

The denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 1 = A(t + 1) + Bt……(ii)

Put t = 0 in above equations we get

1 = A(0 + 1) + B(0)

⇒ A = 1

Now put t = – 1 in equation (ii) we get

1 = A( – 1 + 1) + B( – 1)

⇒ B = – 1

We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = t + 1 ⇒ du = dt, so the above equation becomes,

On integrating we get

Substituting back the values of u, we get

Substituting back the values of t, we get

Applying the logarithm rules, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 24.

Evaluate the following integral:

Answer:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

By equating similar terms, we get

A + C = 0 ⇒ A = – C ………..(iii)

B + D = 0⇒ B = – D…………(iv)

– Ab² – Ca² = 1

⇒ – ( – C)b² – Ca² = 1 (from equation(iii))

…………..(v)

– b²B – a²D = 0

⇒ – b²( – D) – a²D = 0

⇒ D = 0

So equation(iv) becomes B = 0

So equation (iii) becomes,

We put the values of A, B, C, and D values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x² – a²⇒ du = 2dx

v = x² – b²⇒ dv = 2dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Applying the logarithm rule we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 25.

Evaluate the following integral:

Answer:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

By equating similar terms, we get

A + C = 0 ⇒ A = – C ………..(iii)

B + D = 1⇒ B = 1 – D…………(iv)

25A + 4C = 0

⇒ 25( – C) + 4C = 0 (from equation(iii))

⇒ C = 0………..(v)

So equation(iv) becomes

So equation (iii) becomes, A = 0

We put the values of A, B, C, and D values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

in first partthe

in second parthe t

so the above equation becomes,

On integrating we get

(the standard integral of )

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

Question 26.

Evaluate the following integral:

Answer:

Let

Now,

Let

2x + 1 = A(x – 1) + B(x + 1)

Put x = 1

2 + 1 = A × 0 + B × 2

3 = 2B

Put x = – 1

– 2 + 1 = – 2A + B × 0

– 1 = – 2A

Therefore,

Question 27.

Evaluate the following integral:

Answer:

3x – 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)²

Put x = – 1

– 3 – 2 = A × 0 + B × ( – 1 + 3) + C × 0

– 5 = 2B

Put x = – 3

– 9 – 2 = C × ( – 2)( – 2)

– 11 = 4C

Equating coefficients of constants

– 2 = 3A + 3B + C

Thus,

Question 28.

Evaluate the following integral:

Answer:

2x + 1 = A(x – 3)² + B(x + 2)(x – 3) + C(x + 2)

2x + 1 = Ax² – 3Ax + 9A + Bx² – 5Bx – 6B + Cx + 2C

Put x = 3

7 = 5C

Put x = – 2

– 3 = 0A

– 11 = 4C

Equating coefficients of constants

– 2 = 3A + 3B + C

Thus,

Question 29.

Evaluate the following integral:

dx

Answer:

X² + 1 = A(x – 2)(x + 3) + B(x + 3) + C(x – 2)²

Put x = 2

4 + 1 = B × 5

5 = 5B

Put x = – 3

10 = C × 25

Equating coefficients of constants

1 = – 6A + 3B + 4C

Thus,

Question 30.

Evaluate the following integral:

dx

Answer:

x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)²

Put x = – 2

– 2 = 9C

Put x = 1

1 = 3B

Equating coefficients of constants

0 = – 2A + 2B + C

Thus,

Question 31.

Evaluate the following integral:

dx

Answer:

Put x = 1

1 = 4A

Put x = – 1

1 = – 2C

Equating coefficients of x²

1 = A + B

Thus,

Question 32.

Evaluate the following integral:

dx

Answer:

X² + x – 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)²

Put x = – 2

1 = C

Put x = – 1

– 1 = B

Equating coefficients of constants

– 1 = 2A + 2B + C

Thus,

Question 33.

Evaluate the following integral:

Answer:

Equating constants

– 3 = B

Equating coefficients of x

7 = A + 2B

7 = A – 6

A = 13

Equating coefficients of x²

2 = 2A + C

2 = 26 + C

C = – 24

Thus,

Question 34.

Evaluate the following integral:

Answer:

Equating constants

6 = A

Equating coefficients of x²

5 = A + B

B = – 1

Equating coefficients of x

20 = 2A + B + C

20 = 12 – 1 + C

C = 9

Question 35.

Evaluate the following integral:

Answer:

18 = A(x² + 4) + (Bx + C)(x + 2)

Equating constants

18 = 4A + 2C

Equating coefficients of x

0 = 2B + C

Equating coefficients of x²

0 = A + B

Solving, we get

Thus,

Question 36.

Evaluate the following integral:

Answer:

5 = (Ax + B)(x + 2) + C(x² + 1)

Equating constants

5 = 2B + C

Equating coefficients of x

0 = 2A + B

Equating coefficients of x²

0 = A + C

Solving, we get

A = – 1,B = 2,C = 1

Thus

Question 37.

Evaluate the following integral:

Answer:

x = A(x² + 1) + (Bx + C)(x + 1)

Equating constants

0 = A + C

Equating coefficients of x

1 = B + C

Equating coefficients of x²

0 = A + B

Solving, we get

Thus

Question 38.

Evaluate the following integral:

Answer:

1 = (Ax + B)(x + 1) + C(x² + 1)

Equating constants

1 = B + C

Equating coefficients of x

0 = A + B

Equating coefficients of x²

0 = A + C

Solving, we get

Thus

Question 39.

Evaluate the following integral:

Answer:

1 = A(x + 1)(x² + 1) + B(x² + 1) + (Cx + D)(x + 1)²

= Ax³ + Ax² + Ax + A + Bx² + B + Cx³ + 2Cx² + Cx + Dx² + 2D + D

= (A + C)x³ + (A + B + 2C + D)x² + (A + C + 2D)x + (A + B + D)

Equating constants

1 = A + B + D

Equating coefficients of x³

0 = A + C

Equating coefficients of x²

0 = A + B + 2C + D

Equating coefficients of x

0 = A + C + 2D

Solving we get

Thus,

Question 40.

Evaluate the following integral:

Answer:

2x = A(x² + x + 1) + (Bx + C)(x – 1)

= (A + B)x² + (A – B + C)x + (A – C)

Equating constants,

A – C = 0

Equating coefficients of x

2 = A – B + C

Equating coefficients of x²

0 = A + B

On solving,

We get

Question 41.

Evaluate the following integral:

Answer:

1 = (Ax + B)(x² + 4) + (Cx + D)(x² + 1)

= (A + C) x³ + (B + D)x² + (4A + C)x + 4B + D

Equating similar terms

A + C = 0

B + D = 0

4A + C = 0

4B + D = 1

We get,

Thus,

Question 42.

Evaluate the following integral:

Answer:

x² = (Ax + B)(3x² + 4) + (Cx + D)(x² + 1)

= (3A + C) x³ + (3B + D)x² + (4A + C)x + 4B + D

Equating similar terms

3A + C = 0

3B + D = 1

4A + C = 0

4B + D = 0

Solving we get,

A = 0, B = – 1, C = 0,D = 4

Thus,

Question 43.

Evaluate the following integral:

Answer:

3x + 5 = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)²

Put x = 1

8 = 2B

B = 4

Put x = – 1

– 3 + 5 = 4C

2 = 4C

Put x = 0

5 = – A + B + C

Question 44.

Evaluate the following integral:

Answer:

X + 1 = A(x² + 1) + (Bx + C)(x)

Equating constants

A = 1

Equating coefficients of x

1 = C

Equating coefficients of x²

0 = A + B

B = – 1

Question 45.

Evaluate the following integral:

dx

Answer:

X² + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)²

Put x = – 2

3 = C

Put x = – 1

1 = B

Equating coefficients of constants

1 = 2A + 2B + C

Thus,

Question 46.

Evaluate the following integral:

Answer:

Let

1 = A(x⁴ + 1) + (Bx³ + Cx² + Dx + E)(x)

Equating constants

A = 1

Equating coefficients of x⁴

0 = A + B

0 = 1 + B

B = – 1

Equating coefficients of x²

D = 0

Equating coefficients of x

E = 0

Thus,

Question 47.

Evaluate the following integral:

Answer:

Consider the integral,

Rewriting the above integral, we have

Substitute x³ = t

3x²dx = dt

1 = A(t + 8) + Bt

Equating constants

1 = 8A

Equating coefficients of t

0 = A + B

Question 48.

Evaluate the following integral:

Answer:

3 = A(1 + x²) + (Bx + C)(1 – x)

Equating similar terms

A – B = 0

B – C = 0

A + C = 3

Solving

Thus,

Question 49.

Evaluate the following integral:

Answer:

Let

Sin x = t

Cos x dx = dt

1 = A(1 – t)²(2 + t) + B(1 – t)(2 + t) + C(2 + t) + D(1 – t)³

Put t = 1

1 = 3C

Put t = – 2

1 = 27D

Put t = sin x

Question 50.

Evaluate the following integral:

Answer:

Put x² = t

2xdx = dt

2t + 1 = A(t + 4) + Bt

Equating constants

1 = 4A

Equating coefficients of t

2 = A + B

Thus we have

Question 51.

Evaluate the following integral:

Answer:

We have,

Let 1 – sin x = t

⇒ – cos x dx = dt

⇒ I= – (ln t – ln(1 + t)) + c

⇒ I= ln (1 + t) – ln t + c

Question 52.

Evaluate the following integral:

Answer:

⇒ 2x + 1=A(x – 3) + B(x – 2)

⇒ 2x + 1=(A + B)x – 3A – 2B

Equating similar terms, we get,

A + B=2 and 3A + 2B= – 1

So, A= – 5, B=7

⇒ I = – 5 log |x – 2| + 7 log |x – 3| + c

⇒ I = log |x – 2| ^{– 5} + log |x – 3|⁷ + c

Question 53.

Evaluate the following integral:

Answer:

Let, x²=y

⇒ 1=A(y + 2) + B(y + 1)

⇒ 1=(A + B)y + 2A + B

On equating similar terms, we get,

A + B=0, and 2A + B=1

We get, A=1, B= – 1

Question 54.

Evaluate the following integral:

Answer:

⇒ 1= A(x + 1)(x – 1)(x² + 1) + Bx(x – 1)(x² + 1) + cx(x + 1)(x² + 1) + Dx(x + 1)(x – 1)

For, x=0, A= – 1

Question 55.

Evaluate the following integral:

Answer:

⇒ 1= A(x – 1)(x² + 1) + B(x + 1)(x² + 1) + c(x + 1)(x – 1)

Question 56.

Evaluate the following integral:

Answer:

Let x² + 2 = t ⇒ 2x dx = dt

⇒ 1 = At^{2 +} B t (t – 1) + C(t – 1)

For t=1, A=1

For t=0, C= – 1

For t= – 1, B= – 1

Question 57.

Evaluate the following integral:

Answer:

⇒ x² = A(x² + 1) + B(x – 1)

Question 58.

Evaluate the following integral:

Answer:

Let x² = y

⇒ y = A(y + b²) + B(y + a²)

⇒ y = y(A + B) + (Ab² + Ba²)

Equating the coefficients, we get,

A + B=1, and Ab² + Ba² = 0

Question 59.

Evaluate the following integral:

Answer:

Multiplying and dividing by cos x

Let, sin x = t, cos x dx = dt

Question 60.

Evaluate the following integral:

Answer:

Multiplying and dividing by sin x

Let cos x = t, – sin x dx = dt

⇒ 1 = A(t + 1)(3 + 2t) + B(t – 1)(3 + 2t) + C(t² – 1)

Question 61.

Evaluate the following integral:

Answer:

Multiplying and dividing by sin x

Let cos x = t, – sin x dx = dt

Question 62.

Evaluate the following integral:

Answer:

Question 63.

Evaluate the following integral:

Answer:

For, x=0, 10 = 4B + 3D .... (i)

For, x=1, 14 = 5A + 5B + 4C + 4D .... (ii)

For, x= – 1,14 = – 5A + 5B – 4C + 4D .... (iii)

Also, by comparing coefficient of x³ we get, 0=A + C (iv)

On solving, (i), (ii), (iii), (iv) we get,

A=0, B= – 2, C=0, D=6

Question 64.

Evaluate the following integral:

Answer:

Let x²=y

⇒ 4y² + 3 = A(y + 3)(y + 4) + B(y + 2)(y + 4) + C(y + 2)(y + 3)

Question 65.

Evaluate the following integral:

Answer:

Question 66.

Evaluate the following integral:

Answer:

⇒ x² = A(x + 2)(x² + 3) + B(x – 2)(x² + 3) + C(x – 2)(x + 2)

Question 67.

Evaluate the following integral:

Answer:

⇒ x²= A(1 + x)(x² + 1) + B(1 – x)(x² + 1) + c(x + 1)(1 – x)

Question 68.

Evaluate the following integral:

Answer:

⇒ x²=A(x – 1)(x² + 2) + B(x + 1)(x² + 2) + C(x² – 1)

Question 69.

Evaluate the following integral:

Answer:

For, x=0, 19 = – 5B + 3D .... (i)

For, x=1, 26 = – 4A – 4B + 4C + 4D .... (ii)

For, x= – 1,14 = 4A – 4B – 4C + 4D .... (iii)

Also, by comparing coefficient of x³ we get, 0=A + C (iv)

On solving, (i), (ii), (iii), (iv) we get,

---

Exercise 19.4

Question 1.

Evaluate:

Answer:

By doing long division of the given equation we get

Quotient = x + 3

Remainder = –4

∴ We can write the above equation as

⇒ x + 3

∴ The above equation becomes

⇒

⇒

We know

⇒ +c. (Where c is some arbitrary constant)

Question 2.

Evaluate:

Answer:

By doing long division of the given equation we get

Quotient = x²+2x+4

Remainder = 8

∴ We can write the above equation as

⇒ x²+2x+4

∴ The above equation becomes

⇒

⇒

We know

⇒

⇒ . (Where c is some arbitrary constant)

Question 3.

Evaluate:

Answer:

By doing long division of the given equation we get

Quotient =

Remainder =

∴ We can write the above equation as

⇒

∴ The above equation becomes

⇒

⇒

We know

⇒

⇒ . (Where c is some arbitrary constant)

Question 4.

Evaluate:

Answer:

The above equation can be written as

⇒

⇒

⇒

We know

⇒

⇒

⇒ + c. (Where c is an arbitrary constant)

Question 5.

Evaluate:

Answer:

⇒

⇒

⇒

⇒

⇒

⇒

We know

⇒

⇒

⇒ (Where c is some arbitrary constant)

Question 6.

Evaluate:

Answer:

In this question degree of denominator is larger than that of numerator so we need to manipulate numerator.

⇒

⇒

⇒

We know

⇒

⇒ (where c is some arbitrary constant)

---

Exercise 19.31

Question 1.

Evaluate the following integral:

Answer:

re-writing the given equation as

Let as t

Using identity

Substituting t as

Question 2.

Evaluate the following integral:

Answer:

let as

re-writing the given equation as

Let and

So and

Using identity and

Substituting t as and z as

Question 3.

Evaluate the following integral:

Answer:

re-writing the given equation as

Let

Using identity

Substituting t as

Question 4.

Evaluate the following integral:

Answer:

re-writing the given equation as

Let and

and

Using identity and

Substituting t as and z as

Question 5.

Evaluate the following integral:

Answer:

re-writing the given equation as

Substituting t as and z as

and

Using identity

Substituting t as and z as x²

Question 6.

Evaluate the following integral:

Answer:

re-writing the given equation as

Substituting t as

Using identity

Substituting t as

Question 7.

Evaluate the following integral:

Answer:

re-writing the given equation as

Assume

Using identity

Substituting t as

Question 8.

Evaluate the following integral:

Answer:

re-writing the given equation as

Assume

Using identity

Substituting t as

Question 9.

Evaluate the following integral:

Answer:

re-writing the given equation as

Substituting t as and z as

and

Using identity

Substituting t as and z as

Question 10.

Evaluate the following integral:

Answer:

re-writing the given equation as

Assume and

and

Using identity

Substituting t as and z as

Question 11.

Evaluate the following integral:

Answer:

Re-writing the given equation as

Multiplying sec⁴x in both numerator and denominator

=

Assume tanx = t

sec²xdx=dt

=

=

=

Assume

⇒

=

Using identity

=

=

=

---

Exercise 19.32

Question 1.

Evaluate the following integral:

Answer:

assume x+2=t²

dx=2tdt

Using identity

Question 2.

Evaluate the following integral:

Answer:

assume 2x+3=t²

dx=tdt

Using identity

Question 3.

Evaluate the following integral:

Answer:

re-writing the given equation as

Now splitting the integral in two parts

For the first part using identity

For the second part

assume x+2=t²

dx=2tdt

Using identity

Hence integral is

Question 4.

Evaluate the following integral:

Answer:

re-writing the given equation as

For the first- and second-part using identity

For the second part

assume x+2=t²

dx=2tdt

Using identity

Hence integral is

Question 5.

Evaluate the following integral:

Answer:

re-writing the given equation as

For the first part using identity

For the second part

assume x+1=t²

dx=2tdt

Using identity

Hence integral is

Question 6.

Evaluate the following integral:

Answer:

let x=t²

dx=2tdt

Dividing by t² in both numerator and denominator

Let and

and

Using identity and

Substituting and

Question 7.

Evaluate the following integral:

Answer:

assume x+1=t²

dx=2tdt

Dividing by t² in both numerator and denominator

Let

Using identity

Substituting

Substituting

Question 8.

Evaluate the following integral:

Answer:

assume

Using identity

Substituting

Question 9.

Evaluate the following integral:

Answer:

assume

Using identity

Substituting

Question 10.

Evaluate the following integral:

Answer:

assume

Let 1+t²=u²

tdt=udu

Using identity

Substituting

Substituting

Question 11.

Evaluate the following integral:

Answer:

assume x²+1=u²

xdx=udu

Using identity

Substituting

Question 12.

Evaluate the following integral:

Answer:

assume

Let t² - 1=u²

tdt=udu

Using identity

Substituting

Substituting

Question 13.

Evaluate the following integral:

Answer:

assume

Assume 1-4t²=u²

-4tdt=udu

Using identity

Substituting

Substituting

Question 14.

Evaluate the following integral:

Answer:

assume x²+9=u²

xdx=udu

Using identity

Substituting

---

Very Short Answer

Question 1.

Write a value of

Answer:

let x + log sin x= t

Differentiating it on both sides we get,

(1+cot x) dx=dt - i

Given that

Substituting i in above equation we get,

=log t + c

= log(x + log sin x ) + c

Question 2.

Write a value of

Answer:

Consider ∫ e^{3 logx} x⁴

= x³

∫ e^{3 logx} x⁴ = ∫ x³x⁴dx

= ∫ x⁷ dx

Question 3.

Write a value of

Answer:

let x³ = t

Differentiating on both sides we get,

3 x² dx = dt

substituting above equation in we get,

Question 4.

Write a value of

Answer:

let tan x = t

Differentiating on both sides we get,

sec² x dx = dt

Substituting above equation in ∫tan³x sec²x dx we get,

= ∫t³ dt

Question 5.

Write a value of

Answer:

we know ∫e^x (f(x) +f^’(x))dx〗= e^x f(x)+c

Given, ∫ e^x (sin x + cos x ) dx

Here and

Therefore

Question 6.

Write a value of

Answer:

let tan x=t

Differentiating on both sides we get,

sec²x dx = dt

Substituting above equation in ∫ tan³x sec²x dx we get,

=∫t⁶ dt

Question 7.

Write a value of

Answer:

let 3+2sin x=t

Differentiating on both sides we get,

2cos x dx=dt

Substituting above equation in we get,

Question 8.

Write a value of

Answer:

given,

∫ e^x sec x( 1 + tan x) dx = ∫ e^x (sec x + sec x tan x) dx

= e^x sec x + c

∵∫e^x (f(x) +f'(x)) dx=e^xf(x)+c

Question 9.

Write a value of

Answer:

let log xⁿ =t

Differentiating on both sides we get,

Substituting above equations in we get,

Question 10.

Write a value of

Answer:

let log x=t

Differentiating on both sides we get,

Substituting above equations in we get,

Question 11.

Write a value of

Answer:

given ∫e^{log sin x} cos x dx

=∫sin x cos x dx (∵e^logx =x)

Let

Differentiating on both sides we get,

Cos x dx=dt

Substituting above equations in given equation we get,

=∫t dt

Question 12.

Write a value of

Answer:

let sin x=t

Differentiating on both sides we get,

Cos x dx=dt

Substituting above equation in ∫sin³ x cos x dx we get,

=∫t³ dt

Question 13.

Write a value of

Answer:

let cos x=t

Differentiating on both sides we get,

-sin x dx=dt

Substituting above equation in ∫cos⁴ x sin x dx we get,

=∫-t⁴ dt

Question 14.

Write a value of

Answer:

given ∫ tan x sec³ x dx

= ∫ (tan x sec x ) sec² x dx

Let sec x=t

Differentiating on both sides we get,

tan x sec x dx=dt

Substituting above equation in ∫ tan x sec³ x dx we get,

=∫t² dt

Question 15.

Write a value of

Answer:

given

Let 1+e^x =t

Differentiating on both sides we get,

E^x dx=dt

Substituting above equation in given equation we get,

=t- log t + c

=1+e^x -log(1+e^x )+c

Question 16.

Write a value of

Answer:

Take e^x out from the denominator.

Let, e^-x + 2 = t

Differentiating both sides with respect to x

-dt = e^-x dx

Use formula

Y = -ln t + c

Again, put e^-x + 2 = t

Y = -ln(e^-x + 2) + c

Note: Don’t forget to replace t with the function of x at the end of solution. Always put constant c with indefinite integral.

Question 17.

Write a value of

Answer:

Let, tan^-1x = t

Differentiating both sides with respect to x

y= ∫t³ dt

Use formula

Again, put t = tan^-1x

Question 18.

Write a value of

Answer:

Let, tan x = t

Differentiating both side with respect to x

⇒ dt = sec²x dx

Use formula

Again, put t = tan x

Question 19.

Write a value of

Answer:

We know that

1 + sin2x = sin²x + cos²x + 2sinxcosx = (sin x + cos x)²

y= ∫dx

Use formula ∫c dx=cx, where c is constant

y = x + c

Question 20.

Write a value of

Answer:

By using integration by parts

Let, log_e x as Ist function and 1 as IInd function

Use formula

y=x log_e x- ∫dx

y=x log_e x -x + c

Question 21.

Write a value of

Answer:

We know that a and e are constant so, a^x e^x = (ae)^x

Use formula where c is constant

Question 22.

Write a value of

Answer:

We know that e^a+b = e^ae^b

Let, x² = t

Differentiating both sides with respect to x

Use formula

Again, put t = x²

Question 23.

Write a value of

Answer:

We know that by using property of logarithm

y= ∫a^x + x^a dx〗

y= ∫ a^x dx+ ∫ x^a dx

Use formula

Question 24.

Write a value of

Answer:

Let log(sin x) = t

Differentiating both sides with respect to x

Use formula

y = log t + c

Again, put t = log(sin x)

y = log(log(sin x)) + c

Question 25.

Write a value of

Answer:

We know that cos²x = 1 – sin²x

(a²sin²x + b²cos²x) = a²sin²x + b²(1 – sin²x)

= (a² – b²)sin²x + b²

Let, sin²x = t

Differentiating both sides with respect to x

= sin 2x

⇒ dt = sin2x dx

Use formula

Again, put t = sin²x

Question 26.

Write a value of

Answer:

Let, 3 + a^x = t

Differentiating both sides with respect to x

Use formula

Again, put t = 3 + a^x

Question 27.

Write a value of

Answer:

Let, x(log x) = t

Differentiating both sides with respect to x

⇒ dt = (1 + log x)dx

Use formula

y = log(3 + t) + c

Again, put t = x(log x)

y = log(3 + x(log x)) + c

Question 28.

Write a value of

Answer:

Let, cos x = t

Differentiating both sides with respect to x

⇒ -dt = sin x dx

Use formula

Again, put t = cos x

Question 29.

Write a value of

Answer:

We know that

1 + sin2x = sin²x + cos²x + 2sinxcosx

= (sin x + cos x)²

Let, sin x + cos x = t

Differentiating both sides with respect to x

-dt = (sin x – cos x)dx

Use formula

y = -log t + c

Again, put t = sin x + cos x

y = -log(sin x + cos x) + c

Question 30.

Write a value of

Answer:

Let, log x = t

Differentiating both sides with respect to x

Use formula

Again, put t = log x

Question 31.

Write a value of

Answer:

we know ∫f(x)g(x)= f(x) ∫g(x)-∫f' (x) ∫g(x)

Let

Given that

Question 32.

Write a value of s

Answer:

we know ∫f(x)g(x)= f(x) ∫g(x)-∫f' (x) ∫g(x)

Let

Given that

Question 33.

Write a value of

Answer:

given

Question 34.

Write a value of

Answer:

given

Question 35.

Write a value of

Answer:

we know that

Given

Question 36.

Write a value of

Answer:

we know that |

Given

Question 37.

Write a value of

Answer:

we know that

Given

Question 38.

Evaluate:

Answer:

let

Differentiating on both sides we get,

substituting it in we get,

Question 39.

Evaluate:

Answer:

let

Differentiating on both sides we get,

Substituting it in we get,

Question 40.

Evaluate:

Answer:

let

Differentiating on both sides we get,

substituting it in we get,

=2 tan t+c

Question 41.

Evaluate:

Answer:

let

Differentiating on both sides we get,

substituting it in we get,

=∫2 sin t dt

=-2 cos t+c

Question 42.

Evaluate:

Answer:

let

Differentiating on both sides we get,

substituting it in we get,

=∫2cos t dt〗

=2 sin t+c

Question 43.

Evaluate:

Answer:

let 1 + log x = t

Differentiating on both sides we get,

Substituting it in we get,

=∫t² dt

Question 44.

Evaluate:

Answer:

let 7 – 4x = t

Differentiating on both sides we get,

-4 dx = dt

substituting it in we get,

=tan t+c

=tan (7-4x)+c

Question 45.

Evaluate:

Answer:

given

=∫log x

=x log x-x + c

Question 46.

Evaluate:

Answer:

Given,

[since, ]

Question 47.

Evaluate:

Answer:

Given,

=∫sec²x-tanx.sec x dx[since,]

= tan x-sec x + c

Question 48.

Evaluate:

Answer:

Given,

[since,= ]

Question 49.

Evaluate:

Answer:

Given,

=∫(x² +1)dx [since, = ]

Question 50.

Evaluate:

Answer:

Given,

Let tan^-1x=t

ð dt

ð

Now,

=

= +c

= +c

Question 51.

Evaluate:

Answer:

Given,

=sin^-1x + c

(It is a standard formula).

Question 52.

Evaluate:

Answer:

Given, ∫sec x (sec x + tan x ) dx

=∫ (sec² x + sec x. tan x) dx

= tan x + sec x + c

Question 53.

Evaluate:

Answer:

Given,

We know that,

By comparison, a=4

Question 54.

Evaluate:

Answer:

Given, dx

[since,= ]

Question 55.

Evaluate:

Answer:

Given,

_{Let 3x}² +sin6x =t

=dt

⇒ 6x + cos 6x. 6=dt

Substituting the values,

=

= log t +c

= log(3x² + sin6x) +c

Question 56.

If then write the value of f(x).

Answer:

Consider, dx

= e^x dx

= e^x dx

It is clearly of the form,

By comparison, f(x)= ; f^I(x)= -

Therefore, the value of f(x)=

Question 57.

If then write the value f(x).

Answer:

Given,

It is clearly of the form,

By comparison, f(x)=1+ ; f^I(x)=

= e^x (1+tanx) +C

Therefore, the value of f(x)=1+tanx

Question 58.

Evaluate:

Answer:

Given,

_{We Know that, cos2x=1-2sin}²x

⇒ 1-cos2x=2sin²x

Substitute this in the given,

= dx

= dx

= ∫cosec² x dx

= -cotx +c

Question 59.

Write the anti-derivative of

Answer:

Anti-derivative is nothing but integration

_{Therefore its Anti-derivative can be found by integrating the above given equation.}

Question 60.

Evaluate:

Answer:

Given, ∫cos^-1(sin x ) dx

Let us consider, ∫cos^-1dx

We know that, f(x).g(x) dx=g(x) dx-[f^I(x)] dx

By comparison, f(x) =cos^-1x ; g(x)=1

( since, )

=x cos^-1x – (1-x²)^1/2 +c

Therefore,

Replace ‘x’ with :-

ð

=sinx.cos^-1x (sinx) –cosx+c

Question 61.

Evaluate:

Answer:

Given,

= [since, sin²x+cos²x=1]

=

= dx

=)dx

= +c

Question 62.

Evaluate:

Answer:

Given,

Let 1+log x=t

=logt +c

=log (1+logx)+c

---

Mcq

Question 1.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D.

Answer:

-

=

∵ {∫e^x f(x)+f^’x ]=e^x f(x)}

Question 2.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D.

Answer:

-

=

∵ {∫e^x f(x)+f^’x ]=e^x f(x)}

Question 3.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D.

Answer:

⇒ cos x=t then ;

⇒-sin (x)dx=dt

(

put

Question 4.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D.

Answer:

⇒ cos x=t then ;

⇒-sin (x)dx=dt

(

put

Question 5.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D.

Answer:

{=

Question 6.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D.

Answer:

{=

Question 7.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D.

Answer:

Given

if t=e^2x +1

;then

Question 8.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D.

Answer:

Given

if t=e^2x +1

;then

Question 9.

Mark the correct alternative in each of the following:

Evaluate

A. 2 log_e cos (xe^x) + C

B. sec (xe^x) + C

C. tan (xe^x) + C

D. tan (x + e^x) +C

Answer:

let (t)=x;

= tan t

(put (t)= x)

= tan (xe^x) + c

Question 10.

Mark the correct alternative in each of the following:

Evaluate

A. 2 log_e cos (xe^x) + C

B. sec (xe^x) + C

C. tan (xe^x) + C

D. tan (x + e^x) +C

Answer:

let (t)=x;

= tan t

(put (t)= x)

= tan (xe^x) + c

Question 11.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D. none of these

Answer:

I =

⇒ tanx =t

+c

Question 12.

Mark the correct alternative in each of the following:

Evaluate

A.

B.

C.

D. none of these

Answer:

I =

⇒ tanx =t

+c

Question 13.

Mark the correct alternative in each of the following:

The primitive of the function is

A.

B.

C.

D.

Answer:

I=d

=∫a^t dt

Question 14.

Mark the correct alternative in each of the following:

The primitive of the function is

A.

B.

C.

D.

Answer:

I=d

=∫a^t dt

Question 15.

Mark the correct alternative in each of the following:

The value of is

A. 1 + logx

B. x + logx

C. x log(1 + logx)

D. log (1 + logx)

Answer:

I = d

⇒let(1+log_e x)=t

⇒I=log(1+log x)+C

Question 16.

Mark the correct alternative in each of the following:

The value of is

A. 1 + logx

B. x + logx

C. x log(1 + logx)

D. log (1 + logx)

Answer:

I = d

⇒let(1+log_e x)=t

⇒I=log(1+log x)+C

Question 17.

Mark the correct alternative in each of the following:

is equal to

A.

B.

C.

D.

Answer:

let x= dx=2)

I= ∫ (sin t)² dt

I=∫(1-cos 2t)dt

I=∫1dt -∫cos 2t dt

Question 18.

Mark the correct alternative in each of the following:

is equal to

A.

B.

C.

D.

Answer:

let x= dx=2)

I= ∫ (sin t)² dt

I=∫(1-cos 2t)dt

I=∫1dt -∫cos 2t dt

Question 19.

Mark the correct alternative in each of the following:



A. e^x f(x) + C

B. e^x + f(x) + C

C. 2e^x f(x) + C

D. e^x – f(x) + C

Answer:

let I=d

Open the brackets, we get

I={∫ e^x f(x) dx + ∫ e^x f^’(x) dx}

=U+∫ e^x f^’(x) dx

U=∫e^x f(x)dx

To solve U using integration by parts

U = f(x) ∫e^x dx - ∫[f^’(x) ∫e^x]

= f(x) e^x -∫ f^’(x) e^x

= U + ∫ e^x f^’(x) dx

I = e^x f(x) + ∫f^’(x) e^x dx - ∫ e^x f^’(x) dx

I=e^x f(x)+c

Question 20.

Mark the correct alternative in each of the following:



A. e^x f(x) + C

B. e^x + f(x) + C

C. 2e^x f(x) + C

D. e^x – f(x) + C

Answer:

let I=d

Open the brackets, we get

I={∫ e^x f(x) dx + ∫ e^x f^’(x) dx}

=U+∫ e^x f^’(x) dx

U=∫e^x f(x)dx

To solve U using integration by parts

U = f(x) ∫e^x dx - ∫[f^’(x) ∫e^x]

= f(x) e^x -∫ f^’(x) e^x

= U + ∫ e^x f^’(x) dx

I = e^x f(x) + ∫f^’(x) e^x dx - ∫ e^x f^’(x) dx

I=e^x f(x)+c

Question 21.

Mark the correct alternative in each of the following:

The value of is equal to

A.

B.

C. ± (sinx – cosx) + C

D. ±log (sinx – cosx) +C

Answer:

I=

Let t=sin x-cos x

I=±log(sin x-cos x)+c

Question 22.

Mark the correct alternative in each of the following:

The value of is equal to

A.

B.

C. ± (sinx – cosx) + C

D. ±log (sinx – cosx) +C

Answer:

I=

Let t=sin x-cos x

I=±log(sin x-cos x)+c

Question 23.

Mark the correct alternative in each of the following:

If then α is equal to

A. sin x + C

B. cos x + C

C. C

D. none of these

Answer:

using integration by parts

I=∫x sin x d〗

I = x cos x + ∫ cos x dx

(∵ ∫sin x=-cos x)

= x cos x + sin x + c

Question 24.

Mark the correct alternative in each of the following:

If then α is equal to

A. sin x + C

B. cos x + C

C. C

D. none of these

Answer:

using integration by parts

I=∫x sin x d〗

I = x cos x + ∫ cos x dx

(∵ ∫sin x=-cos x)

= x cos x + sin x + c

Question 25.

Mark the correct alternative in each of the following:



A. tan x – x + C

B. x + tan x + C

C. x – tan x + C

D. –x – cot x + C

Answer:

I = - ∫ (tan x )² dx

I = - ∫ (-1 + (sec x)² dx

= (x-tan x) + c

Question 26.

Mark the correct alternative in each of the following:



A. tan x – x + C

B. x + tan x + C

C. x – tan x + C

D. –x – cot x + C

Answer:

I = - ∫ (tan x )² dx

I = - ∫ (-1 + (sec x)² dx

= (x-tan x) + c

Question 27.

Mark the correct alternative in each of the following:

is equal to

A. 2(sinx + x cosθ) + C

B. 2(sinx – x cosθ) + C

C. 2(sinx + 2x cosθ) + C

D. 2(sinx – 2x cosθ) + C

Answer:

I= d

I=2∫(cos x+ cos θ) dx

I 2(sin x+x cos θ) + c

Question 28.

Mark the correct alternative in each of the following:

is equal to

A. 2(sinx + x cosθ) + C

B. 2(sinx – x cosθ) + C

C. 2(sinx + 2x cosθ) + C

D. 2(sinx – 2x cosθ) + C

Answer:

I= d

I=2∫(cos x+ cos θ) dx

I 2(sin x+x cos θ) + c

Question 29.

Mark the correct alternative in each of the following:

is equal to

A.

B.

C.

D.

Answer:

I=

Let

Question 30.

Mark the correct alternative in each of the following:

is equal to

A.

B.

C.

D.

Answer:

I=

Let

Question 31.

Mark the correct alternative in each of the following:

then

A.

B.

C.

D.

Answer:

let =t

[a= ]; [b=-1]

Question 32.

Mark the correct alternative in each of the following:

then

A.

B.

C.

D.

Answer:

let =t

[a= ]; [b=-1]

Question 33.

Mark the correct alternative in each of the following:



A.

B.

C.

D.

Answer:

Question 34.

Mark the correct alternative in each of the following:



A.

B.

C.

D.

Answer:

Question 35.

Mark the correct alternative in each of the following:

If a log |1 + x² +b tan^–1

then

A.

B.

C.

D.

Answer:

= (compare coefficient of both side)

put the value of A,B,C in U

Question 36.

Mark the correct alternative in each of the following:

If a log |1 + x² +b tan^–1

then

A.

B.

C.

D.

Answer:

= (compare coefficient of both side)

put the value of A,B,C in U

---

Revision Exercise

Question 1.

Evaluate

Answer:

Rationalising denominator

We get,

It becomes

= dx

Question 2.

Evaluate

Answer:

Rationalising denominator

We get,

It becomes

= dx

Question 3.

Evaluate

Answer:

Factorising the equation

On cancelling we get

=∫(1+x)(1+x²)dx

=∫(1+x+x² +x³)dx

Question 4.

Evaluate

Answer:

Factorising the equation

On cancelling we get

=∫(1+x)(1+x²)dx

=∫(1+x+x² +x³)dx

Question 5.

Evaluate

Answer:

On simplifying we get,

dx

On solving we get

Question 6.

Evaluate

Answer:

On simplifying we get,

dx

On solving we get

Question 7.

Evaluate

Answer:

On simplifying we get,

Question 8.

Evaluate

Answer:

On simplifying we get,

Question 9.

Evaluate

Answer:

On simplifying we get

= -x^-1 + ln(1+x) + c

Question 10.

Evaluate

Answer:

On simplifying we get

= -x^-1 + ln(1+x) + c

Question 11.

Evaluate

Answer:

On squaring numerator we get

Formula for

Solving above equation we get,

Question 12.

Evaluate

Answer:

On squaring numerator we get

Formula for

Solving above equation we get,

Question 13.

Evaluate

Answer:

Multiplying numerator and denominator with 1-sinx

We get

Taking and

Solving for A

Taking cos x=t

On differentiating both sides we get

-sin x dx=dt

Putting values in A we get our equation as

= t^-1+c

Substituting value of t,

=sec x + c

Solving for B

= ∫ sec²x - ∫ 1 dx

= tan x – x +c

Final answer is A+B

= sec x + tan x – x + c

Question 14.

Evaluate

Answer:

Multiplying numerator and denominator with 1-sinx

We get

Taking and

Solving for A

Taking cos x=t

On differentiating both sides we get

-sin x dx=dt

Putting values in A we get our equation as

= t^-1+c

Substituting value of t,

=sec x + c

Solving for B

= ∫ sec²x - ∫ 1 dx

= tan x – x +c

Final answer is A+B

= sec x + tan x – x + c

Question 15.

Evaluate

Answer:

On simplifying we get

Question 16.

Evaluate

Answer:

On simplifying we get

Question 17.

Evaluate

Answer:

∫ sec²x (cos²x – sin²x )² dx

Opening the square

On multiplying equation reduces to

=∫(cos²x-2sin²x+sec²x-2+cos²x)dx

=∫(2cos²x-2sin²x+sec²x-2)dx

=∫(2(cos²x-sin²x)+sec²x-2)dx

=∫(2cos2x+sec²x-2)dx

On solving this we get our answer i.e

=sin2x+tanx-2x+c

Question 18.

Evaluate

Answer:

∫ sec²x (cos²x – sin²x )² dx

Opening the square

On multiplying equation reduces to

=∫(cos²x-2sin²x+sec²x-2+cos²x)dx

=∫(2cos²x-2sin²x+sec²x-2)dx

=∫(2(cos²x-sin²x)+sec²x-2)dx

=∫(2cos2x+sec²x-2)dx

On solving this we get our answer i.e

=sin2x+tanx-2x+c

Question 19.

Evaluate

Answer:

∫cosec²x(cos²x-sin²x)²dx

Opening the square

On multiplying (1-sin²x)(1-sin²x) equation reduces to

=∫(cosec²x-2+sin²x-2cos²x+ sin²x)dx

=∫(cosec²x-2+2sin²x-2cos²x)dx

=∫(-2(cos²x-sin²x)+cosec²x-2)dx

=∫(-2cos2x+cosec²x-2)dx

On solving this we get our answer i.e

=-sin2x-cotx-2x+c

Question 20.

Evaluate

Answer:

∫cosec²x(cos²x-sin²x)²dx

Opening the square

On multiplying (1-sin²x)(1-sin²x) equation reduces to

=∫(cosec²x-2+sin²x-2cos²x+ sin²x)dx

=∫(cosec²x-2+2sin²x-2cos²x)dx

=∫(-2(cos²x-sin²x)+cosec²x-2)dx

=∫(-2cos2x+cosec²x-2)dx

On solving this we get our answer i.e

=-sin2x-cotx-2x+c

Question 21.

Evaluate

Answer:

Replacing 2x by t

We get dx=dt/2 by differentiating both sides

Our equation has become

simplifying sin²t.cos²t

on multiplying and dividing by 4 we get sin²t.cos²tdt as sin²2t

Hence our final answer is

Question 22.

Evaluate

Answer:

Replacing 2x by t

We get dx=dt/2 by differentiating both sides

Our equation has become

simplifying sin²t.cos²t

on multiplying and dividing by 4 we get sin²t.cos²tdt as sin²2t

Hence our final answer is

Question 23.

Evaluate

Answer:

We can write ∫cos³3xdx as:

∫cos3x(cos²3x)dx and

further as:

=cos3x(1-sin²3x)dx

=∫cos3xdx-∫cos3x(sin²3x)dx

Taking A=∫cos3xdx

Solving for A

A=

Taking B=∫cos3x(sin²3x)dx

In this taking sin3x=t

Differentiating on both sides we get

3cos3xdx=dt

Solving by putting these values we get

B

Substituting values we get

B

Our final answer is A+B i.e

Question 24.

Evaluate

Answer:

We can write ∫cos³3xdx as:

∫cos3x(cos²3x)dx and

further as:

=cos3x(1-sin²3x)dx

=∫cos3xdx-∫cos3x(sin²3x)dx

Taking A=∫cos3xdx

Solving for A

A=

Taking B=∫cos3x(sin²3x)dx

In this taking sin3x=t

Differentiating on both sides we get

3cos3xdx=dt

Solving by putting these values we get

B

Substituting values we get

B

Our final answer is A+B i.e

Question 25.

Evaluate

Answer:

Taking b² common, we get,

taking

on differentiating both sides we get

2sinxcosxdx=dt

Means sin2xdx=dt

putting and sin2xdx=dt in equation we get our equation as

On solving this we get

Substituting value of t we get our answer as

Question 26.

Evaluate

Answer:

Taking b² common, we get,

taking

on differentiating both sides we get

2sinxcosxdx=dt

Means sin2xdx=dt

putting and sin2xdx=dt in equation we get our equation as

On solving this we get

Substituting value of t we get our answer as

Question 27.

Evaluate

Answer:

Taking sin^-1x=t

Differentiating both sides,

We get

Our new equation has become

On solving this we get

Substituting value of t= sin^-1x

We get our answer as

=ln(sin^-1x )+c

Question 28.

Evaluate

Answer:

Taking sin^-1x=t

Differentiating both sides,

We get

Our new equation has become

On solving this we get

Substituting value of t= sin^-1x

We get our answer as

=ln(sin^-1x )+c

Question 29.

Evaluate

Answer:

Taking sin^-1x=t

Differentiating both sides,

We get

Our new equation has become

∫t³dt

On solving this we get

Substituting value of t= sin^-1x

We get our answer as

Question 30.

Evaluate

Answer:

Taking sin^-1x=t

Differentiating both sides,

We get

Our new equation has become

∫t³dt

On solving this we get

Substituting value of t= sin^-1x

We get our answer as

Question 31.

Evaluate

Answer:

We can write above integral as

Considering first integral:

Since the numerator and denominator are exactly same, our integrand simplifies to 1 and integrand becomes:

⇒ ∫ dx

⇒ x

--- (3)

Considering second integral:

Let u = 1 + e^x, du = e^xdx

Apply u – substitution:

Replacing the value of u we get,

---(4)

From (3) and (4) we get,

Question 32.

Evaluate

Answer:

We can write above integrand as:

Considering integrand (A)

Put e^x+1 = t

Differentiating w.r.t x we get,

e^xdx = dt

Substituting values we get

Substituting the value of t we get,

--(i)

Considering integrand (B)

We can write above integral as

(1) (2)

Considering first integral:

Since the numerator and denominator are exactly same, our integrand simplifies to 1 and integrand becomes:

⇒ ∫ dx

⇒ x

--- (3)

Considering second integral:

Let u = 1 + e^x, du = e^xdx

Apply u – substitution:

Replacing the value of u we get,

---(4)

From (3) and (4) we get,

--(ii)

From (i) and (ii) we get,

Question 33.

Evaluate

Answer:

We can write above integral as:

--(1)

Let e^x = t

Differentiating w.r.t x we get,

e^x dx = dt

∴ integral (1) becomes,

= tan^-1(t) + C

Putting value of t we get,

= tan^-1(e^x) + C

Question 34.

Evaluate

Answer:

We can write above integral as:

--(1)

Put Sinx = t

Differentiting w.r.t x we get,

Cosx.dx = dt

∴ integral (1) becomes,

-- (∵ sin²(x) + cos²(x) = 1)

Putting value of t = Sin(x) we get,

Question 35.

Evaluate

Answer:

We can write above integral as:

--(1)

We know that,

2 sinA.sinB = cos(A-B) – cos(A+B)

Now, considering A as x and B as 2x we get,

= 2 sinx.sin2x = cos(x-2x) – cos(x+2x)

= 2 sinx.sin2x = cos(-x) – cos(3x)

= 2 sinx.sin2x = cos(x) – cos(3x) [∵ cos(-x) = cos(x)]

∴ integral (1) becomes,

Cosidering

We know,

2 sinA.cosB = sin(A+B) + sin(A-B)

Now, considering A as 3x and B as x we get,

2 sin3x.cosx = sin(4x) + sin(2x)

--(2)

Again, Cosidering

We know,

2 sinA.cosB = sin(A+B) + sin(A-B)

Now, considering A as 3x and B as 3x we get,

2 sin3x.cos3x = sin(6x) + sin(0)

= sin(6x)

--(3)

∴ integral becomes,

[From (2) and (3)]

Question 36.

Evaluate

Answer:

We can write above integral as:

--(1)

We know that,

2 cosA.cosB = cos(A+B) + cos(A-B)

Now, considering A as x and B as 2x we get,

= 2 cosx.cos2x = cos(x+2x) + cos(x-2x)

= 2 cosx.cos2x = cos(3x) + cos(-x)

= 2 cosx.cos2x = cos(3x) + cos(x) [∵ cos(-x) = cos(x)]

∴ integral (1) becomes,

Cosidering

We know,

2 cosA.cosB = cos(A+B) + cos(A-B)

Now, considering A as x and B as 3x we get,

2 cosx.cos3x = cos(4x) + cos(-2x)

2 cosx.cos3x = cos(4x) + cos(2x) [∵ cos(-x) = cos(x)]

--(2)

Cosidering ∫ 2cos²3x

We know,

cos2A = 2cos²A – 1

2cos²A = 1 + cos2A

Now, considering A as 3x we get,

∫ 2cos²3x = ∫ 1 + cos2(3x) = ∫ 1 + cos(6x)

--(3)

∴ integral becomes,

[From (2) and (3)]

Question 37.

Answer:

We can write above integral as

[Adding and subtracting 1 in denominator]

∵ sin²x + cos²x = 1 and

sin2x = 2 sinx cosx

∵ sin²x + cos²x - 2 sinx cosx = (sinx – cosx)²

Put sinx – cosx = t

Differentiating w.r.t x we get,

(cosx + sinx)dx = dt

Putting values we get,

Putting value of t we get,

Question 38.

Answer:

We can write above integral as

[Adding and subtracting 1 in denominator]

∵ sin²x + cos²x = 1 and

sin2x = 2 sinx cosx

∵sin²x + cos²x + 2 sinx cosx = (sinx + cosx)²

Taking minus (-) common from numerator we get,

Put sinx + cosx = t

Differentiating w.r.t x we get,

(cosx - sinx)dx = dt

Putting values we get,

We know that,

Here x = t and a = 1

Putting value of t we get,

∴ from (1) we get,

Question 39.

Evaluate

Answer:

Multiply and divide in R.H.S we get,

We can write above integral as:

[∵ sin(A+B) = sinA.cosB – cosA.sinB]

By simplifying we get,

[∵ ∫ cotx dx = log|sinx| + C]

Question 40.

Evaluate

Answer:

Multiply and divide in R.H.S we get,

We can write above integral as:

[∵ sin(A+B) = sinA.cosB – cosA.sinB]

By simplifying we get,

[∵ ∫ tanx dx = -log|cosx| + C]

Question 41.

Evaluate

Answer:

We can write above integral as:

(Adding and subtracting 1 in numerator)

Consider

(∵ sin²x + cos²x = 1 and sin2x = 2 sinx.cosx)

--- (1)

[From (1)]

Considering,

(Adding and subtracting 1 in denominator)

--(2)

Putting values in (2) we get,

Substituting value of u we get,

Question 42.

Evaluate

Answer:

We know cos 2x = 2cos²x – 1

Putting values in I we get,

Put cosx = t

Differentiating w.r.t to x we get,

sinx dx = -dt

Putting values in integral we get,

Again put √2× t = u

Differentiating w.r.t to t we get,

Putting values in integral we get,

Substituting value of u we get,

Substituting value of t we get,

Question 43.

Evaluate

Answer:

We can write above integral as:

----(Splitting tan³x)

(Using tan²x = sec²x – 1)

Considering integral (1)

Let u = tanx

du = sec²x dx

Substituting values we get,

Substituting value of u we get,

∴ integral becomes,

[∵ ∫ tanx dx = -log|cosx| + C]

Question 44.

Answer:

We can write above integral as:

----(Splitting tan⁴x)

(Using tan²x = sec²x – 1)

Considering integral (1)

Let u = tanx

du = sec²x dx

Substituting values we get,

Substituting value of u we get,

Considering integral (2)

∴ integral becomes,

[∵ C+C is a constant]

Question 45.

Answer:

We can write above integral as:

----(Splitting tan⁵x)

(Using tan²x = sec²x – 1)

----(Splitting tan³x)

(Using tan²x = sec²x – 1)

Considering integral (1)

Let u = tanx

du = sec²x dx

Substituting values we get,

Substituting value of u we get,

Considering integral (2)

Let t = tanx

dt = sec²x dx

Substituting values we get,

Substituting value of t we get,

Considering integral (3)

[∵ ∫ tanx dx = -log|cosx| + C]

∴ integral becomes,

[∵ C+C+C is a constant]

Question 46.

Evaluate

Answer:

In this question, first of all we expand cot⁴x as

cot⁴x = (cosec²x – 1)²

= cosec⁴x – 2cosec²x + 1 …(1)

Now, write cosec⁴x as

cosec⁴x = cosec²xcosec²x

= cosec²x(1 + cot²x)

= cosec²x + cosec²xcot²x

Putting the value of cosec⁴x in eq(1)

cot⁴x = cosec²x + cosec²xcot²x – 2cosec²x + 1

= cosec²xcot²x – cosec²x + 1

y = ∫ cot⁴x dx

= ∫ cosec²x cot²x dx + ∫ - cosec²x + 1 dx

A = ∫cosec²x cot²x dx

Let, cot x = t

Differentiating both side with respect to x

⇒ -dt = cosec²x dx

= ∫-t² dt

Using formula

Again, put t = cot x

Now, B= ∫-cosec² x +1 dx

Using formula ∫cosec² x dx= -cot x and ∫c dx=cx

B = cot x + x + c₂

Now, the complete solution is

y = A + B

Question 47.

Evaluate

Answer:

Let, sin x = t

Differentiating both sides with respect to x

⇒ dt = cos x dx

Using formula and

Again, put t = sin x

Question 48.

Evaluate

Answer:

Using formula and

Question 49.

Evaluate

Answer:

In this question we write as

Using formula

Question 50.

Evaluate

Answer:

Let, x = tan t

Differentiating both side with respect to t

⇒ dx = sec²t dt

Again, let cos t = z

Differentiating both side with respect to t

⇒ -dz = sin t dt

Using formula and

Again, put z = cos t = cos(tan^-1x)

Question 51.

Evaluate

Answer:

Let, sin x² = t

Differentiating both sides with respect to x

Using formula

Again, put t = sin x²

Question 52.

Evaluate

Answer:

Let, cos x = t

Differentiating both side with respect to x

⇒ -dt = sin x dx

Using formula

Again, put t = cos x

Question 53.

Evaluate

Answer:

Let, cos x = t

Differentiating both side with respect to x

⇒ -dt = sin x dx

Using formula and

Again, put t = cos x

Question 54.

Evaluate .

Answer:

Let, sin x = t

Differentiating both side with respect to x

⇒ dt = cos x dx

Using formula and

Again, put t = sin x

Question 55.

Evaluate

Answer:

Let, sin x = t

Differentiating both side with respect to x

⇒ dt = cos x dx

Using formula

Again, put t = sin x

Question 56.

Evaluate

Answer:

Let, sin²x = t

Differentiating both side with respect to x

⇒ dt = sin 2x dx

Try to make perfect square in denominator

Using formula

Again, put t = sin²x

Question 57.

Evaluate

Answer:

Let, x = a sec t

Differentiating both side with respect to t

⇒ dx = a sec t tan t dt

Using formula

y = ln(tan t + sec t) + c₁

Again, put

Question 58.

Evaluate

Answer:

Let, x = a tan t

Differentiating both side with respect to t

⇒ dx = a sec²t dt

Tip: This is very important formula. It is use directly in the question. So, learn it by heart.

Using formula

y = ln(tan t + sec t) + c₁

Again, put

Question 59.

Evaluate

Answer:

In this question we can try to make perfect square in

denominator

Using formula

Question 60.

Evaluate

Answer:

In this question we can try to make perfect square in

denominator

Using formula

Question 61.

Evaluate

Answer:

Given,

It is clearly of the form,

Where

Question 62.

Evaluate

Answer:

Given,

Now, 3x²+13x-10

= 3x²+15x-2x-10

= 3x(x+5)-2(x-5)

= (x-5) (3x-2)

1≅ A (3x-2) + B(x+5)

Equating ‘x’ coeff: -

0=3A+B

B=-3A

Equating constant:-

1=-2A+5B

1=-2A+5(-3A)

1=-2A-15A

1=-17A

Question 63.

Evaluate

Answer:

Given,

Let cosx=t

-sinx dx=dt

Now, t²-2t-3

= t²-3t+t-3

= t(t-3)+t-3

= (t-3) (t+1)

1≅ A(t-1)+B(t-3)

Equating ’t’ coeff:-

0=A+B

A=-B

Equating constant:-

1=-A-3B

1=-(-B)-3B

1=-2B

Question 64.

Evaluate

Answer:

Given,dx

Rationalising the denominator:-

Let

Clearly, it is of the form

Where

Question 65.

Evaluate

Answer:

Given,

It is clearly of the form,

Where, a=2; x=x+1

Question 66.

Evaluate

Answer:

Given,

Consider, x+1 A

x+1≅A(2x+4)+B

Equating ‘x’coeff:-

1=2A

equating constant:-

1=4A+B

1=2+B

B=-1

x+1≅ 1/2 (2x+4)-1

Now,

Question 67.

Evaluate

Answer:

Given,

Now,

5x+7 ≅ A (2x-9)+B

_{Equating’x’ coeff:-}

5=2A

A=

_{Equating constant:-}

7=-9A+B

Question 68.

Evaluate

Answer:

Given,

Let

⇒ u² = x+1

⇒ u² -1 = x

2 du = dx

As we know,

Now substitute back the value of u.

Question 69.

Evaluate

Answer:

Given,

Let

_{dx =2t dt}

Now,

Consider, t=sin k

dt=cos k dk

=2∫cos²k dk

=∫2 cos²k dk

=∫cos 2k-1 dk [since, cos 2x=2cos²x-1]

=t cos(sin^-1 t) -2sin^-1 t+2c

=√x cos(sin^-1√x )-2 sin^-1√x+2c

Question 70.

Evaluate

Answer:

Given,

Let

Now,

ax = (1+ t)²

Put back the value of t to get,

Question 71.

Evaluate

Answer:

Given,

Let tanx=t

Sec²x dx=dt

Now,

Now,

1≅ A(t-2)+B(2t+1)

Equating ‘t’ coeff: -

0=A+2B

A=-2B

Equating constant: -

1=-2A+B

1=-2(-2B) +B

_1=5B

B=

Now,

Question 72.

Evaluate

Answer:

Given,

Let tanx=t

Question 73.

Evaluate

Answer:

Given,

Consider, a=b=1

=

=

=

Now, cosx=A (cosx+sinx) +B (cosx+sinx)

=A (cosx+sinx) +B (-sinx+cosx)

Equating ’cosx’ coeff:- Equating ‘sinx’ coeff:-

1=A+B 0=A-B

A=B

1=A+A

2A=1

A=1/2 B=1/2

[since,

(x)+

Question 74.

Evaluate

Answer:

Given,

Let

cosec²x dx=dt

Now,

Question 75.

Evaluate

Answer:

Given,

Equating ‘sin x’ coeff: -

1=2A-B

B=2A-1

Equating ’cos x’ coeff:-

2=A+2B

2=A+2(2A-1)

2=A+4A-2

4=5A

Now,

Question 76.

Evaluate

Answer:

Given,

Put, x⁴=t

4x³dx=dt

x³dx=

Question 77.

Evaluate

Answer:

Given,

Put tanx=t

sec²x dx=dt

dx =

and cos2x=

Now,

Question 78.

Evaluate

Answer:

Given,

Let

[since,

Question 79.

Evaluate

Answer:

Given,

Put

dx = dt and

Question 80.

Evaluate

Answer:

Given,

Let

dx = and

Question 81.

Evaluate

Answer:

To solve this type of solution, we are going to substitute the value of sinx and cosx in terms of tan(x/2)

In this type of equations, we apply substitution method so that equation may be solve in simple way

Let

Put these terms in above equation,we get

Let us now again apply the substitution method in above equation

Let t^-2 = k

-2.t^-3dt = dk

Substitute these terms in above equation gives-

Now put the value of t, t=tan(x/2) in above equation gives us the finally solution

Question 82.

Evaluate

Answer:

To solve this type of solution ,we are going to substitute the value of sinx and cosx in terms of tan(x/2)

In this type of equations we apply substitution method so that equation may be solve in simple way

Let

Put these terms in above equation,we get

Let us now again apply the substitution method in above equation

Let t^-2 = k

-2.t^-3dt = dk

Substitute these terms in above equation gives-

Now put the value of t, t=tan(x/2) in above equation gives us the finally solution

Question 83.

Evaluate

Answer:

Consider ,

Divide num and denominator by cos⁴x to get,

Let tan x = t

sec²x dx = dt

Now divide both numerator and denominator by to get,

Let

Question 84.

Evaluate

Answer:

in this integral we are going to put the value of sin (x) in terms of tan(x/2)-

By applying the formula of 1/(x²+a²) in above equation yields the integral-

By putting the value of t in above equation ,

Question 85.

Evaluate

Answer:

above equation can be solve by using one formula that is (i+ tan²x=sec²x)

_{I = ∫ sec}⁴x dx

_{= ∫ sec}²x sec²x dx

_{= ∫ sec}²x ( 1 + tan²x ) dx

_{= ∫ sec}²x dx + ∫ sec²x tan²x dx

Put tanx=t in above equation so that sec²xdx=dt

Question 86.

Evaluate

Answer:

above equation can we solve by the formula of (1+cot²x=cosec²x)

_{I = ∫ cosec}⁴2x dx

_{= ∫cosec}²2x (1+cot²2x) dx

_{= ∫ cosec}²2x dx + ∫cosec²2x cot²2x dx

Let us consider that cot2x=t then -2.cosec²2xdx=dt

Question 87.

Evaluate

Answer:

first divide nominator by denominator –

: To solve this type of solution ,we are going to substitute the value of sinx and cosx in terms of tan(x/2)

In this type of equations we apply substitution method so that equation may be solve in simple way

Let tan(x/2)=t

1/2.sec²(x/2)dx=dt

Put these terms in above equation,we get

Substitute these terms in above equation gives-

Now put the value of t, t=tan(x/2) in above equation gives us the finally solution

Question 88.

Evaluate

Answer:

To solve this type of solution ,we are going to substitute the value of sinx and cosx in terms of tan(x/2)

In this type of equations we apply substitution method so that equation may be solve in simple way

Let tan(x/2)=t

1/2.sec²(x/2)dx=dt

Put these terms in above equation,we get

Question 89.

Evaluate

Answer:

to solve this integral we have to apply substitution method for which we are going to put x=a.tan²k

_{This means dx = 2.a.tank.sec}²k.dk,then I will be,

In this above integral let tank =t then sec²kdk=dt ,put in above equation-

Apply the formula of sqrt(x²+a²)=x/2.sqrt(a²+x²)+a²/2ln|x+sqrt(a²+x²)|

Now put the value of t in above integral t=tank,then finally integral will be-

Now put the value of k in terms of x that is tan²k=x/a in above integral –

Question 90.

Evaluate

Answer:

Let, 6 + x – 2x² = t

Differentiating both side with respect to t

⇒ (1 – 4x)dx = dt

Again, put t = 6 + x – 2x²

Make perfect square of quadratic equation

6 + x – 2x²

Use formula

The final solution of the question is y = A + B

Question 91.

Evaluate

Answer:

to solve this type of integration we have to let cosx either sinx =t then manuplate them

_{Let cos x =t then –sinx dx =dt}

_{Also apply the formula of (sin}²t+cos²t=1)

Now put the value of t in above integral

Question 92.

Evaluate

Answer:

to solve this type of integration we have to let cosx either sinx =t then manuplate them

_{Let sin x =t then cosx dx =dt}

_{Also apply the formula of (sin}²t+cos²t=1)

Now put the value of t in above integral

Question 93.

Evaluate

Answer:

Let, sin x = t

Differentiating both side with respect to t

⇒ cos x dx = dt

Again put t = sin x

Question 94.

Evaluate

Answer:

dividing by cos⁶x yields-

I=∫tan²x.sec⁴x dx

Let us consider tanx=t

Then sec²xdx=dt,put in above equation-

Now reput the value of t,which is t=tanx

Question 95.

Evaluate

Answer:

in this integral we will use the formula 1+tan²x=sec²x,

_{I = ∫sec}²x sec⁴x dx

_{= ∫sec}²x (1 + tan²x)² dx

Now put tan x=t which means sec²xdx=dt,

I = ∫(1+t²)² dt

= ∫(1+t⁴+2t²) dt

Now put the value of t, which is t=tan x in above integral-

Question 96.

Evaluate

Answer:

in this integral we will use the formula 1+tan²x=sec²x,

_{Then equation will be transform in below form-}

_{I = ∫tan}⁵x sec²x sec x dx

_{= ∫sec x tan}⁵x sec²x dx

Now put tan x=t which means sec²xdx=dt,

In this above integral put 1+t²=k²

^{that is mean tdt=kdk}

I = ∫(k⁴ + 1- 2k) k² dk

= ∫ (k⁶ + k² – 2k³)dk

Now put the value of k=(1+t²)=sec²x in above equation-

Question 97.

Evaluate

Answer:

in this integral we will use the formula 1+tan²x=sec²x,

_{Then equation will be transform in below form-}

_{I = ∫tan}³x sec²x sec²x dx

_{= ∫tan}³x (1 + tan²x) sec²x dx

Now put tanx=t which means sec²xdx=dt,

Now put the value of t,which is t=tanx in above integral-

Question 98.

Evaluate

Answer:

Use 1 = sin²x + cos²x

Use sin x + cos x

Question 99.

Evaluate

Answer:

in these type of problems we put the value of x=a tank

That is mean that dx=a sec²k dk

= ∫ a. sec k. a. sec²k dk

=∫ a² sec³k dk

By upper solve questions we can find out the value of integration of sec³x,whixh is equal to

Put the value of integration of sec³x in above equation we get our finally integral which is –

Now put the value of k which is tan^-1(x/a) in above equation-

Question 100.

Evaluate

Answer:

Consider ,

Let and II = 1

As ∫ I.II dx = I.∫ II dx - ∫ [ d/dx(I). ∫ II dx]

So,

Question 101.

Evaluate

Answer:

Let, x = a sin t

Differentiate both side with respect to t

⇒ dx = a cos t dt

y = ∫a² cos²t dt

Again, put

Question 102.

Evaluate

Answer:

Make perfect square of quadratic equation

3x² + 4x + 1

Using formula,

Question 103.

Evaluate

Answer:

Make perfect square of quadratic equation

1 + 2x – 3x²

Using formula,

Question 104.

Evaluate

Answer:

Make perfect square of quadratic equation

1 + x – x²

Let,

Differentiate both side with respect to t

⇒ dx = dt

Let, t² = z

Differentiate both side with respect to z

Put z = t² and

Put

The final answer is y = A + B

Question 105.

Evaluate

Answer:

Make perfect square of quadratic equation

4x² + 5x + 6

Let,

Differentiate both side with respect to t

⇒ dx = dt

Let, t² = z

Differentiate both side with respect to z

Put z = t² and

Put

+

The final answer is y = A + B

Question 106.

Evaluate

Answer:

Use the method of integration by parts

The final answer is y = A + B

Question 107.

Evaluate

Answer:

Use the method of integration by parts

Question 108.

Evaluate

Answer:

Let, log x = t

Differentiating both side with respect to t

Note:- Always use direct formula for ∫log x dx

y = ∫log t dt

y = t log t – t + c

Again, put t = log x

y = (log x)log(log x) – log x + c

Question 109.

Evaluate

Answer:

Use method of integration by parts

Use formula ∫tan x dx = log secx

Question 110.

Evaluate

Answer:

We know that

Use method of integration by parts

Question 111.

Evaluate

Answer:

y = ∫(x² + 2x + 1) e^x dx

y = ∫(x² + 2x)e^x dx + ∫e^x dx

We know that ∫(f(x) +f’(x))e^x dx = f(x) e^x

Here, f(x) = x² then f’(x) = 2x

y = x²e^x + e^x + c

y = (x² + 1)e^x + c

Question 112.

Evaluate

Answer:

Use method of integration by parts

Let, x² + a² = t

Differentiating both side with respect to t

Again, put t = x² + a²

Question 113.

Evaluate

Answer:

Use method of integration by parts

Question 114.

Evaluate

Answer:

Use method of integration by parts

Question 115.

Evaluate

Answer:

Use method of integration by parts

Question 116.

Evaluate

Answer:

Let,

Differentiate both side with respect to t

Use formula

Again put

Question 117.

Evaluate

Answer:

Let, x = sin²t

Differentiate both side with respect to t

⇒ dx = 2 sin t cos t dt

Let, cos t = z

Differentiate both side with respect to z

⇒ sin t dt = -dz

Again put z = cos t and

Question 118.

Evaluate

Answer:

Let, 1 + x³ = t

Differentiate both side with respect to t

Again, put t = 1 + x³

Question 119.

Evaluate

Answer:

Use formula

Question 120.

Evaluate

Answer:

Let, x = sin t

Differentiate both side with respect to t

⇒ dx = cos t dt

Again put t = sin^-1x

Question 121.

Answer:

Let

Differentiate both side with respect to t

Again, put

Question 122.

Evaluate

Answer:

Let sinx – cosx=t ,

(cosx+sinx)dx=dt

Question 123.

Evaluate

Answer:

Here we will use integration by parts,

Choose u in these oder LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL)

So here,u=tan^-1x

……….(

Putting 1+x² =t,

2xdx=dt

Resubstituting t

Question 124.

Evaluate

Answer:

∫u. dv=uv-∫v du

Choose u in these odder

LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL)

Here u=tan^-1 and v=1.

Put

dx=2tdt

and x=t²

Question 125.

Evaluate

Answer:

Choose u in these order LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL)

u=sin^-1 v=1

Put

dx=2tdt

Now put t=sinu;

dt=cos u du;

=cos u

…(Here we can substitute sin²x=(1-cos2u)/2)

Put

I=

Question 126.

Evaluate

Answer:

Choose u in these order LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL)

Here u= and v=1.

Put x-1=t dx=dt

Question 127.

Evaluate

Answer:

Put x=cos2t;dx=-2sin2t

dx=

Question 128.

Evaluate

Answer:

Put x=atan²t;dx=2a.tant.sec²t dt

Question 129.

Evaluate

Answer:

Put x=sint ;dx=costdt

By by parts,

=3[t sin t-∫sin t dt]+c

=3[t sin t + cos t]+c

Question 130.

Evaluate

Answer:

_{Put x=sin t;}

_{dx=cos t dt}

Question 131.

Evaluate

Answer:

Put x=sin t

_{;dx=cos t dt;}

=+c

Question 132.

Evaluate

Answer:

we can put sin^-1x=t;dx/(1-x²)^1/2=dt;(1-x²)=cos²t and x=sint.

By by parts,

…….

=t sec t-log (tan t + sec t) + C'

Put cost=u;

-sin t dt=du

=-(-u^-1 )+c

=sec t + C

Question 133.

Evaluate

Answer:

Put 2x=t dx=dt/2

Question 134.

Evaluate

Answer:

=

Question 135.

Evaluate

Answer:

…….()

Question 136.

Evaluate

Answer:

Put tan^-1x=t,dx/(1+x²)=dt, 1+x²=sec²x;

Question 137.

Evaluate

Answer:

By using partial differentiation,

By substituting the x² coefficients and other coefficients we can get,

A=-1/8;B=1/8;C=3/4;D=1/2;

Question 138.

Evaluate

Answer:

I₁

put x²+x+1=t;

(2x+1)dx=dt

I₁

Now, I₂

put (2x+1)/ =u;

2dx/=dt;

dx=dt/2

So, answer is ]+c

Question 139.

Evaluate

Answer:

We can write the integral as follows,

Question 140.

Evaluate

Answer:

By partial fractions,

Solving these two equations,2A+5B=1 and A+B=0

We get A=-1/3 and B=1/3

Question 141.

Answer:

_{By partial fractions,}

So by solving, A=- � ;B=2; C=- � ;D = -3/2

Question 142.

Evaluate

Answer:

Let, x = sin²t

Differentiating both side with respect to t

Again, put

Question 143.

Answer:

by partial fraction,

So we get these three equations ,

2A+2B+C=1

3A+B+2C=1

A+C=1

So the values are A=-2;C=3;B=1

Question 144.

Answer:

Put 2x=t;

2dx=dt;dx=dt/2

Question 145.

Evaluate

Answer:

Put cot x=t, -cosec²x dx = dt;

By partial fractions it’s a remembering thing

That if you see the above integral just apply the below return result,

---

Exercise 19.5

Question 1.

Evaluate:

Answer:

In these questions, little manipulation makes the questions easier to solve

Here multiply and divide by 2 we get

⇒

Add and subtract 1 from the numerator

⇒

⇒

⇒

⇒

⇒

⇒

Question 2.

Evaluate:

Answer:

Here Add and subtract 2 from x

We get

⇒

⇒

⇒

Question 3.

Evaluate:

Answer:

In these questions, little manipulation makes the questions easier to solve

Add and subtract 5 from the numerator

⇒

⇒

⇒

⇒

⇒

⇒

Question 4.

Evaluate:

Answer:

Here multiply and divide the question by 3

We get

⇒

⇒

Add and subtract 1 from above equation

⇒

⇒

⇒

⇒

⇒

Question 5.

Evaluate:

Answer:

Let 2x + 1 = λ(3x + 2) + μ

2x + 1 = 3xλ + 2λ + μ

comparing coefficients we get

3λ = 2 ; 2λ + μ = 1

⇒

Replacing 2x + 1 by λ(3x + 2) + μ in the given equation we get

⇒

⇒

⇒

⇒

⇒

Question 6.

Evaluate:

Answer:

Let 3x + 5 = λ(7x + 9) + μ

3x + 5 = 7xλ + 9λ + μ

comparing coefficients, we get

7λ = 3 ; 9λ + μ = 1

⇒

Replacing 3x + 5 by λ(7x + 9) + μ in the given equation we get

⇒

⇒

⇒

⇒

⇒

Question 7.

Evaluate:

Answer:

In these questions, little manipulation makes the questions easier to solve

Add and subtract 4 from the numerator

⇒

⇒

⇒

⇒

⇒

⇒

Question 8.

Evaluate:

Answer:

Let 2 – 3x = λ(3x + 1) + μ

2 – 3x = 3xλ + λ + μ

comparing coefficients we get

3λ = – 3 ; λ + μ = 2

⇒

Replacing 2 – 3x by λ(3x + 1) + μ in given equation we get

⇒

⇒

⇒

⇒

⇒

Question 9.

Evaluate:

Answer:

Let 5x + 3 = λ(2x – 1) + μ

5x + 3 = 2xλ – λ + μ

comparing coefficients we get

2λ = 5 ; – λ + μ = 3

⇒

Replacing 5x + 3 by λ(2x – 1) + μ in the given equation we get

⇒

⇒

⇒

⇒

⇒

Question 10.

Evaluate:

Answer:

Rationalise the given equation we get

⇒

⇒

⇒

⇒

Assume x = √t

⇒

Substituting t and dt

⇒

⇒

⇒

⇒

But x = √t

⇒

---

Exercise 19.6

Question 1.

Evaluate:

∫ sin²(2x + 5) dx

Answer:

sin²x =

∴ The given equation becomes,

⇒

We know

⇒

⇒

Question 2.

Evaluate:

∫ sin³(2x + 1) dx

Answer:

We know sin3x = –4sin³x+3sinx

⇒ 4sin³x = 3sinx–sin3x

⇒

⇒

⇒ We know

⇒

⇒ .

Question 3.

Evaluate:

∫ cos⁴ 2x dx

Answer:

Cos⁴2x = (cos²2x)²

^⇒ cos²x =

⇒

⇒

^⇒ cos²4x =

⇒

Now the question becomes

⇒

We know

⇒

⇒

Question 4.

Evaluate:

∫ sin² b x dx

Answer:

sin²x =

∴ The given equation becomes,

⇒

We know

⇒

⇒

Question 5.

Evaluate:

Answer:

sin²x =

∴ The given equation becomes,

⇒

We know

⇒

⇒

Question 6.

Evaluate:

Answer:

We know,cos²x =

∴ The given equation becomes,

⇒

We know

⇒

⇒

Question 7.

Evaluate:

∫ cos²nx dx

Answer:

We know,cos²x =

∴ The given equation becomes,

⇒

We know

⇒

⇒

Question 8.

Evaluate:

Answer:

⇒ 2sin²x =

We can substitute the above result in the given equation

∴ The given equation becomes

⇒

⇒

sin²x =

⇒

⇒

⇒

---

Exercise 19.7

Question 1.

∫ sin 4x cos 7x dx

Answer:

We know 2sinAcosB = sin(A + B) + sin(A – B)

∴ sin4xcos7x =

We know sin( – θ) = – sinθ

∴ sin( – 3x) = – sin3x

∴ The above equation becomes

⇒

⇒

We know

⇒

⇒ + c

Question 2.

∫ cos 3x cos 4x dx

Answer:

We know 2cosAcosB = cos(A – B) + cos(A + B)

∴ cos4xcos3x =

∴ The above equation becomes

⇒

⇒

We know

⇒

⇒ c

Question 3.

∫ cos mx cos nx dx, m ≠ n

Answer:

We know 2cosAcosB = cos(A – B) + cos(A + B)

∴ cosmxcosnx =

∴ The above equation becomes

⇒

We know

⇒

⇒

Question 4.

∫ sin mx cos nx dx, m ≠ n

Answer:

We know 2sinAcosB = sin(A + B) + sin(A – B)

∴ sinmxcosnx =

∴ The above equation becomes

⇒

We know

⇒

⇒

Question 5.

∫ sin 2x sin 4x sin 6x dx

Answer:

We need to simplify the given equation to make it easier to solve

We know 2sinAsinB = cos(A – B) – cos(A + B)

∴ sin4xsin2x =

∴ The above equation becomes

⇒

⇒

We know 2sinAcosB = sin(A + B) + sin(A – B)

∴ sin6xcos2x =

_{Also 2sinx.cosx = sin2x}

∴ sin6xcos6x =

∴ The above equation simplifies to

⇒

⇒

We know

⇒

⇒

⇒ (where c is some arbitrary constant)

Question 6.

∫ sin x cos 2x sin 3x dx

Answer:

We know 2sinAcosB = sin(A + B) + sin(A – B)

∴ sin3xcos2x =

∴ The given equation becomes

⇒

⇒

We know 2sinAsinB = cos(A – B) – cos(A + B)

∴ sin5xsinx =

Also sin²x =

∴ Above equation can be written as

⇒

⇒

We know

⇒

⇒

⇒

NOTE: – Whenever you are solving integral questions having trigonometric functions in the product then the first thing that should be done is convert them in the form of addition or subtraction.

---

Exercise 19.8

Question 1.

Evaluate the following integrals:

Answer:

In the given equation cos 2x = cos² x – sin² x

Also we know cos² x + sin² x = 1.

∴Substituting the values in the above equation we get

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Question 2.

Evaluate the following integrals:

Answer:

In the given equation

cosx =

Also, = 1

Substituting in the above equation we get,

⇒

⇒

⇒

⇒

⇒

⇒

Question 3.

Evaluate the following integrals:

Answer:

1 + cos 2x = 2 cos²x

1 – cos 2x = 2 sin²x

(both of them are trigonometric formuales)

⇒

⇒

⇒

⇒ ln|sinx| + c

Question 4.

Evaluate the following integrals:

Answer:

1 – cosx =

1 + cosx =

⇒

⇒

⇒

⇒

Question 5.

Evaluate the following integrals:

Answer:

Here first of all convert secx in terms of cosx

∴ We get

⇒

∴ We get

⇒

=

∴ The equation now becomes

⇒

We know

Cos 2x = 2 cos²x - 1

∴ We can write the above equation as

⇒

⇒

⇒ 2 sin x –

(

⇒ 2 sin x – ln|sec x + tan x| + c

Question 6.

Evaluate the following integrals:

Answer:

Expanding (cos x + sin x)² = cos²x + sin²x + 2 sinx cosx

We know cos²x + sin²x = 1, 2sinxcosx = sin2x

∴ (cos x + sin x)² = 1 + sin2x

∴ we can write the given equation as

⇒

Assume 1 + sin2x = t

⇒

⇒ 2cos2x dx = dt

∴ cos2xdx =

Substituting these values in the above equation we get

⇒

⇒ + c

substituting t = 1 + 2 sin x in above equation

⇒

Question 7.

Evaluate the following integrals:

Answer:

While solving these types of questions, it is better to eliminate the denominator.

⇒

Add and subtract b in (x - a)

⇒

⇒

Numerator is of the form sin(A + B) = sinAcosB + cosAsinB

Where A = x - b ; B = b - a

⇒

⇒

⇒

⇒

As

⇒ cos(b - a)x + sin(b - a)ln|sin(x - b)|

Question 8.

Evaluate the following integrals:

Answer:

Add and subtract α in the numerator

⇒

⇒

Numerator is of the form sin(A - B) = sinAcosB - cosAsinB

Where A = x + α ; B = 2α

⇒

⇒

⇒

⇒

As

⇒ cos(2α)x + sin(2α )ln|sin(x + α )|

Question 9.

Evaluate the following integrals:

Answer:

Convert tanx in form of sinx and cosx.

⇒

∴ The equation now becomes

⇒

⇒

⇒

Let cosx - sinx = t

∴

⇒ - (cosx + sinx)dx =dt

Substituting dt and t

We get

⇒

⇒ - ln t + c

t = cosx – sinx

∴ - ln|cosx – sinx| + c

Question 10.

Evaluate the following integrals:

Answer:

Add and subtract a from x in the numerator

∴ The equation becomes

⇒

Numerator is of the form cos(A + B)= cosAcosB - sinAsinB

Where A= x - a ; B= a

⇒

⇒

As

⇒ xcosa – sina + c

Question 11.

Evaluate the following integrals:

Answer:

We know cos²x + sin²x = 1.

Also, 2sinxcosx = sin2x

1 + sin2x =cos²x + sin²x + 2sinxcosx = (cosx + sinx)²

1 - sin2x =cos²x + sin²x - 2sinxcosx = (cosx - sinx)²

∴ The equation becomes

⇒

⇒

Assume cosx + sinx = t

∴ d(cosx + sinx) = dt

= cosx - sinx

∴ dt = cosx – sinx

⇒

= ln|t| + c

But t = cosx + sinx

∴ ln|cosx + sinx| + c.

Question 12.

Evaluate the following integrals:

Answer:

Assume e^3x + 1 = t

⇒ d(e^3x + 1) = dt

⇒ 3e^3x=dt

⇒ e^3x =

Substituting t and dt in the given equation we get

⇒

⇒

⇒

But t = e^3x + 1

∴ The above equation becomes

⇒ .

Question 13.

Evaluate the following integrals:

Answer:

Assume 3secx + 5=t

d(3secx + 5)=dt

3secxtanx=dt

Secxtanx=

Substitute t and dt

We get

⇒

⇒

But t = 3secx + 5

∴ the equation becomes

⇒ .

Question 14.

Evaluate the following integrals:

Answer:

Convert cotx in form of sinx and cosx.

⇒

∴ The equation now becomes

⇒

⇒

⇒

Assume cosx + sinx = t

∴ d(cosx + sinx) = dt

= cosx - sinx

∴ dt = cosx – sinx

⇒

= ln|t| + c

But t = cosx + sinx

∴ ln|cosx + sinx| + c.

Question 15.

Evaluate the following integrals:

Answer:

Assume log(tanx) = t

d(log(tanx)) = dt

⇒

⇒ secx.cosecx.dx=dt

Put t and dt in given equation we get

⇒

= ln|t| + c.

But t = log(tanx)

= ln|log(tanx)| + c.

Question 16.

Evaluate the following integrals:

Answer:

Assume 3 + logx = t

d(3 + logx) =dt

⇒

Put t and dt in given equation we get

⇒

= ln|t| + c.

But t = 3 + logx

= ln|3 + logx| + c

Question 17.

Evaluate the following integrals:

Answer:

Assume e^x + x =t

d(e^x + x)=dt

e^x + 1 = dt

Put t and dt in given equation we get

⇒

= ln|t| + c.

But t = e^x + x

= ln| e^x + 1| + c

Question 18.

Evaluate the following integrals:

Answer:

Assume logx =t

d(logx)=dt

= dt

Put t and dt in given equation we get

⇒

= ln|t| + c.

But t = logx

= ln| logx| + c

Question 19.

Evaluate the following integrals:

Answer:

Assume acos²x + bsin²x = t

d(acos²x + bsin²x) = dt

( - 2acosx.sinx + 2bsinx.cosx)dx = dt

(bsin2x - asin2x)dx=dt

(b - a)sin2xdx=dt

Sin2xdx =

Put t and dt in given equation we get

⇒

= ln|t| + c.

But t = acos²x + bsin²x

= ln| acos²x + bsin²x | + c.

Question 20.

Evaluate the following integrals:

Answer:

Assume 2 + 3sinx = t

d(2 + 3sinx)=dt

3cosxdx = dt

cosxdx=

Put t and dt in given equation we get

⇒

=

But t =2 + 3sinx

=

Question 21.

Evaluate the following integrals:

Answer:

Assume x + cosx = t

d(x + cosx) =dt

⇒ 1 - sinx dx =dt

Put t and dt in given equation we get

⇒

= ln|t| + c.

But t = x + cosx

= ln| x + cosx | + c

Question 22.

Evaluate the following integrals:

Answer:

First of all take e^x common from denominator so we get

⇒

⇒

Assume be ^{- x} + c =t

d(be ^{- x} + c) = dt

⇒ - be ^{- x}dx= dt

⇒

Substituting t and dt we get

⇒

⇒

But t =(be ^{- x} + c)

⇒

Question 23.

Evaluate the following integrals:

Answer:

First of all, take e^x common from the denominator, so we get

⇒

⇒

Assume e ^{- x} + 1 =t

d(e ^{- x} + 1) = dt

⇒ - e ^{- x}dx= dt

⇒ e ^{- x}dx= - dt

Substituting t and dt we get

⇒

⇒

But t =(e ^{- x} + 1)

⇒ .

Question 24.

Evaluate the following integrals:

Answer:

Assume log(sinx)= t

d(log(sinx)) =dt

⇒ dx =dt

⇒ cotx dx = dt

Put t and dt in given equation we get

⇒

= ln|t| + c.

But t = log(sinx)

= ln| log(sinx) | + c

Question 25.

Evaluate the following integrals:

Answer:

Assume e^2x - 2= t

d(e^2x - 2) =dt

⇒ 2e^2xdx =dt

⇒ e^2xdx =

Put t and dt in the given equation we get

⇒

=

But t = e^2x - 2

=

Question 26.

Evaluate the following integrals:

Answer:

Taking 2 common in denominator we get

⇒

Now assume

3cosx + 2sinx = t

( - 3sinx + 2cosx)dx=dt

Put t and dt in given equation we get

⇒

=

But t = 3cosx + 2sinx

=

Question 27.

Evaluate the following integrals:

Answer:

Assume x² + sin2x + 2x =t

d(x² + sin2x + 2x) =dt

(2x + 2cos2x + 2)dx = dt

2(x + cos2x + 1)dx = dt

(x + cos2x + 1)dx =

Put t and dt in given equation we get

⇒

=

But t = x² + sin2x + 2x

=

Question 28.

Evaluate the following integrals:

Answer:

Let I =

Dividing and multiplying I by sin (a – b) we get,

I =

I =

I =

I =

We know that,

Therefore,

I =

Question 29.

Evaluate the following integrals:

Answer:

Assume 2sinx + cosx =t

d(2sinx + cosx) =dt

(2cosx - sinx)dx= dt

Put t and dt in given equation we get

⇒

=

But t = 2sinx + cosx

= ln|2sinx + cosx| + c.

Question 30.

Evaluate the following integrals:

Answer:

Assume sin4x – sin2x =t

d(sin4x – sin2x) =dt

(cos4x – cos2x)dx= dt

Put t and dt in given equation we get

⇒

=

But t = sin4x – sin2x

= ln| sin4x – sin2x | + c.

Question 31.

Evaluate the following integrals:

Answer:

Assume log(secx + tanx) =t

d(log(secx + tanx)) =dt

(use chain rule to differentiate first differentiate log(secx + tanx) then (secx + tanx)

⇒ =dt

⇒

⇒ secx dx =dt

Put t and dt in the given equation we get

⇒

=

But t = log(secx + tanx)

= ln| log(secx + tanx) | + c.

Question 32.

Evaluate the following integrals:

Answer:

Assume log(tan) =t

d(log()) =dt

(use chain rule to differentiate)

⇒ =dt

⇒

⇒

⇒ cosecx dx =dt

Put t and dt in the given equation we get

⇒

=

But t = log(tan)

= ln| log(tan) | + c.

Question 33.

Evaluate the following integrals:

Answer:

Assume log(logx) =t

d(log(logx)) =dt

(use chain rule to differentiate first)

⇒ =dt

Put t and dt in given equation we get

⇒

=

But t = log(log(x))

= ln| log(log(x)) | + c.

Question 34.

Evaluate the following integrals:

Answer:

Assume 1 + cotx =t

d(1 + cotx) =dt

⇒cosec²x=dt

Put t and dt in given equation we get

⇒

=

But t = 1 + cotx

= ln|1 + cotx| + c.

Question 35.

Evaluate the following integrals:

Answer:

Assume 10^x + x¹⁰ =t

d(10^x + x¹⁰) =dt

a^x = log_ea

⇒ 10x⁹ + 10^xlog_e10=dt

Put t and dt in given equation we get

⇒

=

But t = 10^x + x¹⁰

= ln|10^x + x¹⁰| + c.

Question 36.

Evaluate the following integrals:

Answer:

Assume x + cos²x =t

d(x + cos²x) =dt

(1 + ( - 2cosx.sinx))dx = dt

2sinx.cosx=sin2x

(1 - sin2x)dx=dt

Put t and dt in given equation we get

⇒

=

But t = x + cos²x

= ln| x + cos²x | + c.

Question 37.

Evaluate the following integrals:

Answer:

Assume x + logxsecx =t

d(x + logxsecx) =dt

= dt

(1 + tanx)dx = dt

Put t and dt in given equation we get

⇒

=

But t = x + logxsecx

= ln| x + logxsecx | + c.

Question 38.

Evaluate the following integrals:

Answer:

Assume a² + b²sin²x = t

d(a² + b²sin²x) = dt

2b².sinx.cosx.dx=dt

(2sinx.cosx = sin2x)

Sin2xdx =

Put t and dt in the given equation we get

⇒

=

But t = a² + b²sin²x

= .

Question 39.

Evaluate the following integrals:

Answer:

Assume x + logx = t

d(x + logx) = dt

⇒

⇒

Put t and dt in the given equation we get

⇒

=

But t = x + logx

= .

Question 40.

Evaluate the following integrals:

Answer:

Assume 2 + 3sin ^{- 1}x = t

d(2 + 3sin ^{- 1}x) = dt

⇒

⇒

Put t and dt in the given equation we get

⇒

=

But t = 2 + 3sin ^{- 1}x

= .

Question 41.

Evaluate the following integrals:

Answer:

Assume tanx + 2 =t

d(tanx + 2) =dt

(sec²xdx) = dt

Put t and dt in given equation we get

⇒

=

But t = tanx + 2

= ln| tanx + 2 | + c.

Question 42.

Evaluate the following integrals:

Answer:

Assume sin2x + tanx - 5 =t

d(tanx + sin2x - 5) =dt

(2cos2x + sec²x)dx = dt

Put t and dt in given equation we get

⇒

=

But t = sin2x + tanx - 5

= ln|sin2x + tanx - 5| + c.

Question 43.

Evaluate the following integrals:

Answer:

sin2x can be written as sin(5x - 3x)

∴ The equation now becomes

⇒

sin(A - B) = sinAcosB - cosAsinB

⇒

⇒

⇒

⇒

⇒ .

Question 44.

Evaluate the following integrals:

Answer:

Assume x + log(sinx) =t

d(x + log(sinx)) =dt

= dt

(1 + cot)dx = dt

Put t and dt in given equation we get

⇒

=

But t = x + log(sinx)

= ln| x + log(sinx) | + c.

Question 45.

Evaluate the following integrals:

Answer:

Assume √ x + 1 =t

d(√ x + 1) = dt

⇒

⇒

Put t and dt in given equation we get

⇒

=

But t = √ x + 1

=2 ln| √ x + 1 | + c.

Question 46.

Evaluate the following integrals:

∫ tan 2x tan 3x tan 5x dx

Answer:

We know tan5x = tan(2x + 3x)

tan(A + B) =

∴ tan(2x + 3x) =

∴ tan(5x) =

⇒ tan(5x)(1 - tan2x.tan3x) = tan(2x) + tan(3x)

⇒ tan(5x) - tan2x.tan3x.tan5x = tan(2x) + tan(3x)

⇒ tan(5x) - tan(2x) - tan(3x) = tan2x.tan3x.tan5x

Substituting the above result in given equation we get

⇒

⇒

⇒

⇒ .

Question 47.

Evaluate the following integrals:

∫ {1 + tan x tan (x + θ)} dx

Answer:

tan(A - B) =

∴ tan(x - (x + θ )) =

∴ tan(θ) =

⇒ tan(θ)(1 + tanx.tan(x + θ )) = tan(x) - tan(x + θ)

⇒ (1 + tanx.tan(x + θ )) =

⇒

⇒

⇒

⇒

Question 48.

Evaluate the following integrals:

Answer:

sin(A - B) = sinAcosB - cosAsinB

∴ We can write

sin(A + B) = sinAcosB + cosAsinB

∴ We can write

∴ The given equation becomes

⇒

⇒

Denominator is of the form (a - b)(a + b) = a² - b²

⇒ ….(1)

We know sin²x + cos²x = 1

∴ sin²x =1 - cos²x

Substituting the above result in (1) we get

⇒

⇒ …(2)

Let us assume

⇒

⇒ 2sinx.cosx.dx=dt

⇒ sin2x.dx=dt

Substituting dt and t in (2) we get

⇒

= ln|t| + c

But t =

∴ ln| | + c.

Question 49.

Evaluate the following integrals:

Answer:

Multiplying and dividing the numerator by e we get the given as

⇒ …(1)

Assume e^x + x^e = t

⇒ d(e^x + x^e )=dt

⇒ e^x + ex^{e - 1} = dt

Substituting t and dt in equation 1 we get

⇒

= ln|t| + c

But t = e^x + x^e

∴ ln| e^x + x^e | + c.

Question 50.

Evaluate the following integrals:

Answer:

We know sin²x + cos²x = 1

⇒

⇒

⇒

⇒

d(secx) = tanx.secx

∴

∴

∵

⇒ secx + log|| + c.

Question 51.

Evaluate the following integrals:

Answer:

The denominator is of the form cosC - cosD =

∴ cos3x - cosx=

∴ cos3x - cosx= - 2sin2x.sinx

- 2sin2x.sinx = - 2.2.sinx.cosx.sinx

- 2sin2x.sinx= - 4sin²x.cosx

Also sin²x + cos²x = 1

⇒

⇒

⇒

⇒

d(cscx) = cscx.cotx

∴

∴

∵

⇒

---

Exercise 19.9

Question 1.

Evaluate the following integrals:

Answer:

Assume logx = t

⇒ d(logx) = dt

⇒

Substituting t and dt in above equation we get

⇒

⇒

But t = log(x)

⇒ .

Question 2.

Evaluate the following integrals:

Answer:

Assume

⇒

⇒

⇒

⇒

⇒

∴ Substituting t and dt in the given equation we get

⇒

⇒

⇒

But

⇒

Question 3.

Evaluate the following integrals:

Answer:

Assume 1 + √x = t

⇒ d(1 + √x) = dt

⇒

⇒

∴ Substituting t and dt in the given equation we get

⇒

⇒

⇒

But 1 + √x = t

⇒ .

Question 4.

Evaluate the following integrals:

Answer:

Assume 1 + e^x = t

⇒ d(1 + e^x) = dt

⇒ e^xdx = dt

∴ Substituting t and dt in given equation we get

⇒

⇒

⇒

But 1 + e^x = t

⇒ .

Question 5.

Evaluate the following integrals:

Answer:

Assume cosx = t

⇒ d(cos x) = dt

⇒ - sinxdx = dt

⇒ dx =

∴ Substituting t and dt in the given equation we get

⇒

⇒

⇒

But cos x = t

⇒ .

Question 6.

Evaluate the following integrals:

Answer:

Assume 1 + e^x = t

⇒ d(1 + e^x) = dt

⇒ e^xdx = dt

∴ Substituting t and dt in given equation we get

⇒

⇒

⇒

But 1 + e^x = t

⇒ .

Question 7.

Evaluate the following integrals:

∫ cot³x cosec²x dx

Answer:

Assume cotx = t

⇒ d(cotx) = dt

⇒ - cosec²x.dx = dt

⇒

∴ Substituting t and dt in the given equation we get

⇒

⇒

⇒

⇒

But t = cotx

⇒ .

Question 8.

Evaluate the following integrals:

Answer:

Assume sin ^{- 1}x = t

⇒ d(sin ^{- 1}x) = dt

⇒

∴ Substituting t and dt in the given equation we get

⇒

⇒

⇒

But t = sin ^{- 1}x

⇒

Question 9.

Evaluate the following integrals:

Answer:

Assume x - cosx = t

⇒ d(x - cosx) = dt

⇒ (1 + sinx )dx = dt

∴ Substituting t and dt in given equation we get

⇒

⇒

⇒

But t = x – cosx.

⇒ 2(x - cosx)^1/2 + c.

Question 10.

Evaluate the following integrals:

Answer:

Assume sin ^{- 1}x = t

⇒ d(sin ^{- 1}x) = dt

⇒

∴ Substituting t and dt in the given equation we get

⇒

⇒

⇒

But t = sin ^{- 1}x

⇒

Question 11.

Evaluate the following integrals:

Answer:

We know d(sinx) = cosx, and cot can be written in terms of cos and sin

∴

∴ The given equation can be written as

⇒

⇒

Now assume sinx = t

d(sinx) = dt

cosx dx = dt

Substitute values of t and dt in above equation

⇒

⇒

⇒

⇒

⇒

Question 12.

Evaluate the following integrals:

Answer:

We know d(cosx) = sinx, and tan can be written interms of cos and sin

∴

∴ The given equation can be written as

⇒

⇒

Now assume cosx = t

d(cosx) = - dt

sinx dx = - dt

Substitute values of t and dt in above equation

⇒

⇒

⇒

⇒

⇒

Question 13.

Evaluate the following integrals:

Answer:

In this equation, we can manipulate numerator

Cos³x = cos²x.cosx

∴ Now the equation becomes,

⇒

Cos²x = 1 - sin²x

⇒

Now,

Let us assume sinx = t

d(sinx) = dt

cosx dx = dt

Substitute values of t and dt in the above equation

⇒

⇒

⇒

⇒

But t = sinx

⇒ .

Question 14.

Evaluate the following integrals:

Answer:

In this equation, we can manipulate numerator

sin³x = sin²x.sinx

∴ Now the equation becomes,

⇒

sin²x = 1 - cos²x

⇒

Now ,

Let us assume cosx = t

d(cosx) = dt

- sinx dx = dt

Substitute values of t and dt in above equation

⇒

⇒

⇒

⇒

But t = cosx

⇒

Question 15.

Evaluate the following integrals:

Answer:

Assume tan ^{- 1}x = t

d(tan ^{- 1}x) = dt

⇒

Substituting t and dt in above equation we get

⇒

⇒

⇒

But t = tan ^{- 1}x

⇒ 2(tan ^{- 1}x)^1/2 + c.

Question 16.

Evaluate the following integrals:

Answer:

Multiply and divide by cosx

⇒

⇒

⇒

Assume tanx = t

d(tanx) = dt

sec²x dx = dt

Substituting t and dt in above equation we get

⇒

⇒

⇒

But t = tanx

⇒ 2(tanx)^1/2 + c.

Question 17.

Evaluate the following integrals:

Answer:

Assume logx = t

d(log(x)) = dt

⇒

∴ Substituting t and dt in given equation we get

⇒

⇒

⇒

But logx = t

⇒ .

Question 18.

Evaluate the following integrals:

∫ sin⁵ x cos x dx

Answer:

Assume sinx = t

d(sinx) = dt

cosxdx = dt

∴ Substituting t and dt in given equation we get

⇒

⇒

But t = sinx

⇒ .

Question 19.

Evaluate the following integrals:

∫ tan^3/2 x sec² x dx

Answer:

Assume tanx = t

d(tanx) = dt

sec²xdx = dt

∴ Substituting t and dt in given equation we get

⇒

⇒

But t = tanx

⇒

Question 20.

Evaluate the following integrals:

Answer:

Assume x² + 1 = t

⇒d(x² + 1) = dt

⇒2x dx = dt

⇒

x³ can be write as x².x

∴ Now the given equation becomes

⇒

x² + 1 = t ⇒ x² = t - 1

⇒

⇒

⇒

⇒

But t = (x² + 1)

⇒

⇒

⇒

Question 21.

Evaluate the following integrals:

Answer:

Here (4x + 2) can be written as 2(2x + 1).

Now assume, x² + x + 1 = t

d(x² + x + 1) = dt

(2x + 1)dx = dt

⇒

⇒

⇒

⇒

But t = x² + x + 1

⇒ .

Question 22.

Evaluate the following integrals:

Answer:

Assume, 2x² + 3x + 1 = t

d(x² + x + 1) = dt

(4x + 3)dx = dt

Substituting t and dt in above equation we get

⇒

⇒

⇒

But t = 2x² + 3x + 1

⇒ 2(2x² + 3x + 1)^1/2 + c.

Question 23.

Evaluate the following integrals:

Answer:

x = t²

d(x) = 2t.dt

dx = 2t.dt

Substituting t and dt we get

⇒

⇒

Add and subtract 1 from numerator

⇒

⇒

⇒

⇒ 2(t – ln|1 + t|)

But t = √x

⇒ 2(√x – ln|1 + √x|) + c

Question 24.

Evaluate the following integrals:

Answer:

Assume cos²x = t

d(cos²x) = dt

- 2sinxcosxdx = dt

- sin2x.dx = dt

Substituting t and dt

⇒

⇒ e^t + c.

But t = cos²x

⇒ e^cos2x + c

Question 25.

Evaluate the following integrals:

Answer:

Assume x + sinx = t

d(x + sinx) = dt

(1 + cosx)dx = dt

Substituting t and dt in given equation

⇒

⇒

⇒

⇒

But t = x + sinx

⇒

Question 26.

Evaluate the following integrals:

Answer:

We know cos²x + sin²x = 1, 2sinxcosx = sin2x

∴ Denominator can be written as

cos²x + sin²x + 2sinxcosx = (sinx + cosx)²

∴ Now the given equation becomes

⇒

Assume cosx + sinx = t

∴ d(cosx + sinx) = dt

= cosx - sinx

∴ dt = cosx – sinx

⇒

⇒

⇒

⇒

But t = cosx + sinx

⇒

Question 27.

Evaluate the following integrals:

Answer:

Assume a + bcos2x = t

d(a + bcos2x) = dt

- 2bsin2x dx = dt

Sin2xdx =

⇒

⇒

⇒

⇒

But t = a + bcos2x

⇒ .

Question 28.

Evaluate the following integrals:

Answer:

Assume log x = t

⇒ d(logx) = dt

⇒

Substituting the values oft and dt we get

⇒

⇒

But t = logx

⇒ .

Question 29.

Evaluate the following integrals:

Answer:

Assume 1 + cosx = t

⇒ d(1 + cosx) = dt

⇒ - sinx.dx = dt

Substituting the values oft and dt we get

⇒

⇒

⇒

⇒

But t = 1 + cosx

⇒

Question 30.

Evaluate the following integrals:

∫ cotx log sin x dx

Answer:

Assume log(sinx) = t

d(log(sinx)) = dt

⇒

⇒ cot x dx = dt

Substituting the values oft and dt we get

⇒

⇒

But t = log(sinx)

⇒ .

Question 31.

Evaluate the following integrals:

∫ sec x log (sec x + tan x) dx

Answer:

Assume log(secx + tanx) = t

d(log(secx + tanx)) = dt

(use chain rule to differentiate first differentiate log(secx + tanx) then (secx + tanx)

⇒ = dt

⇒

⇒ secx dx = dt

Put t and dt in given equation we get

Substituting the values oft and dt we get

⇒

⇒

But t = log(secx + tanx)

⇒ .

Question 32.

Evaluate the following integrals:

∫ cosec x log (cosec x – cot x) dx

Answer:

Assume log(cosec x – cot x) = t

d(log(cosec x – cot x)) = dt

(use chain rule to differentiate first differentiate log(secx + tanx) then (secx + tanx)

⇒ = dt

⇒

⇒ cscx dx = dt

Put t and dt in given equation we get

Substituting the values oft and dt we get

⇒

⇒

But t = log(cosec x – cot x)

⇒ .

Question 33.

Evaluate the following integrals:

∫ x³ cos x⁴ dx

Answer:

Assume x⁴ = t

d(x⁴) = dt

4x³dx = dt

x³dx =

Substituting t and dt

⇒

⇒

But t = x⁴

⇒ .

Question 34.

Evaluate the following integrals:

∫ x³ sin x⁴ dx

Answer:

Assume x⁴ = t

d(x⁴) = dt

4x³dx = dt

x³dx =

Substituting t and dt

⇒

⇒

But t = x⁴

⇒ .

Question 35.

Evaluate the following integrals:

Answer:

Assume sin ^{- 1}x² = t

⇒ d(sin ^{- 1}x) = dt

⇒

⇒

∴ Substituting t and dt in given equation we get

⇒

⇒

⇒

But t = sin ^{- 1}x

⇒

Question 36.

Evaluate the following integrals:

∫ x³ sin (x⁴ + 1) dx

Answer:

Assume x⁴ + 1 = t

d(x⁴ + 1) = dt

4x³dx = dt

x³dx =

Substituting t and dt

⇒

⇒

But t = x⁴ + 1

⇒ .

Question 37.

Evaluate the following integrals:

Answer:

Assume xe^x = t

d(xe^x) = dt

(e^x + xe^x) dx = dt

e^x(1 + x) dx = dt

Substituting t and dt

⇒

⇒

⇒ tan t + c

But t = xe^x + 1

⇒ tan (xe^x) + c.

Question 38.

Evaluate the following integrals:

Answer:

Assume

⇒

⇒

⇒

Substituting t and dt

⇒

⇒

But t =

⇒

Question 39.

Evaluate the following integrals:

∫ 2x sec³ (x² + 3) tan (x² + 3) dx

Answer:

sec³ (x² + 3) can be written as sec² (x² + 3). sec (x² + 3)

Now the question becomes

⇒

Assume sec (x² + 3) = t

d(sec (x² + 3)) = dt

2x sec (x² + 3) tan (x² + 3)dx = dt

Substituting t and dt in the given equation

⇒

⇒

⇒ .

Question 40.

Evaluate the following integrals:

Answer:

Assume (x + logx) = t

d(x + logx) = dt

⇒

⇒

Substituting t and dt

⇒

⇒

But t = x + logx

⇒

Question 41.

Evaluate the following integrals:

Answer:

Assume 1 - tan²x = t

d(1 - tan²x) = dt

2.tanx.sec²xdx = dt

Substituting t and dt we get

⇒ ⇒

⇒

⇒

But t = 1 - tan²x

⇒ .

Question 42.

Evaluate the following integrals:

Answer:

Assume 1 + (logx)² = t

d(1 + (logx)²) = dt

⇒

⇒

⇒

⇒

⇒

But t = 1 + (logx)²

⇒ .

Question 43.

Evaluate the following integrals:

Answer:

Assume

⇒

Substituting t and dt we get

⇒

^⇒ cos²x =

∴ The given equation becomes,

⇒

We know

⇒

⇒

But

⇒ .

Question 44.

Evaluate the following integrals:

∫ sec⁴ x tan x dx

Answer:

Put tanx = t

d(tanx) = dt

sec²xdx = dt

⇒

We can write sec⁴x = sec²x. sec²x

Now ,the question becomes

⇒

⇒

Tan²x + 1 = sec²x

tanx = t

t² + 1 = sec²x

⇒

⇒

⇒

But t = tanx

⇒

Question 45.

Evaluate the following integrals:

Answer:

Assume e^√x = t

d(e^√x) = dt

⇒

⇒

Substituting t and dt

⇒2

= 2sint + c

But t = e^√x

⇒2 sin(e^√x ) + c.

Question 46.

Evaluate the following integrals:

Answer:

Assume

⇒

Substituting t and dt we get

⇒

^⇒ cos²x =

∴ The given equation becomes,

⇒

We know

⇒

⇒

But

⇒ .

Question 47.

Evaluate the following integrals:

Answer:

Assume √x = t

d(√x) = dt

⇒

⇒

Substituting t and dt

⇒2

= - 2cost + c

But √x = t

⇒2 cos(√x) + c.

Question 48.

Evaluate the following integrals:

Answer:

Assume xe^x = t

d(xe^x) = dt

(e^x + xe^x) dx = dt

e^x(1 + x) dx = dt

Substituting t and dt

⇒

⇒

⇒ - cot t + c

But t = xe^x + 1

⇒ - cot (xe^x) + c.

Question 49.

Evaluate the following integrals:

Answer:

Assume x + tan ^{- 1}x = t

d(x + tan ^{- 1}x) = dt

⇒

⇒

Substituting t and dt

⇒

⇒

But t = x + tan ^{- 1}x

⇒ .

Question 50.

Evaluate the following integrals:

Answer:

Assume sin ^{- 1}x = t

d( sin ^{- 1}x) = dt

⇒

∴ Substituting t and dt in given equation we get

⇒

⇒

But t = sin ^{- 1}x

⇒

Question 51.

Evaluate the following integrals:

Answer:

Assume √x = t

d(√x) = dt

⇒

⇒

Substituting t and dt

⇒2

= 2sint + c

But √x = t

⇒2 sin(√x) + c.

Question 52.

Evaluate the following integrals:

Answer:

Assume tan ^{- 1}x = t

d( tan ^{- 1}x) = dt

⇒

Substituting t and dt

⇒

= - cost + c

But t = tan ^{- 1}x

⇒ - cos(tan ^{- 1}x) + c.

Question 53.

Evaluate the following integrals:

Answer:

Assume logx = t

d(logx) = dt

⇒

Substituting t and dt

⇒

= - cost + c

But t = logx

⇒ cos(logx) + c.

Question 54.

Evaluate the following integrals:

Answer:

Assume tan ^{- 1}x = t

d( tan ^{- 1}x) = dt

⇒

Substituting t and dt

⇒

⇒

But t = tan ^{- 1}x

⇒ .

Question 55.

Evaluate the following integrals:

Answer:

Rationlize the given equation we get

⇒

⇒

Assume x² = t

2x.dx = dt

⇒

Substituting t and dt

⇒

⇒

⇒

⇒

But t = x²

⇒

Question 56.

Evaluate the following integrals:

Answer:

Assume tan ^{- 1}x² = t

d( tan ^{- 1}x²) = dt

⇒

⇒

Substituting t and dt

⇒

⇒

But t = tan ^{- 1}x²

⇒ .

Question 57.

Evaluate the following integrals:

Answer:

Assume sin ^{- 1}x = t

d( sin ^{- 1}x) = dt

⇒

∴ Substituting t and dt in given equation we get

⇒

⇒

But t = sin ^{- 1}x

⇒ .

Question 58.

Evaluate the following integrals:

Answer:

Assume 2 + 3logx = t

d(2 + 3logx) = dt

⇒

⇒

Substituting t and dt

⇒

= - cost + c

But t = 2 + 3logx

⇒

Question 59.

Evaluate the following integrals:

Answer:

Assume x² = t

⇒ 2x.dx = dt

⇒

Substituting t and dt

⇒

⇒

But x² = t

⇒ .

Question 60.

Evaluate the following integrals:

Answer:

Assume 1 + e^x = t

e^x = t - 1

d(1 + e^x) = dt

e^x dx = dt

dx =

Substitute t and dt we get

⇒

⇒

⇒

⇒

⇒

But t = 1 + e^x

⇒

Question 61.

Evaluate the following integrals:

Answer:

Assume √x = t

d(√x) = dt

⇒

⇒

Substituting t and dt

⇒2∫sec²t dt

= 2tant + c

But √x = t

⇒2 tan(√x) + c.

Question 62.

Evaluate the following integrals:

∫ tan³2x sec 2x dx

Answer:

tan³2x. sec 2x = tan²2x. tan2x.sec2x.dx

tan²2x = sec²2x - 1

⇒ tan²2x. tan2x.sec2x.dx = (sec²2x - 1). tan2x.sec2x.dx

⇒ sec²2x tan2x.sec2xdx - tan2x.sec2xdx

∴

⇒

Assume sec2x = t

d(sec2x) = dt

sec2x.tan2x.dx = dt

⇒

⇒

But t = sec2x

⇒ .

Question 63.

Evaluate the following integrals:

Answer:

The given equation can be written as

⇒

First integration be I1 and second be I2.

⇒ For I1

Add and subtract 2 from the numerator

⇒

⇒

⇒

⇒ x - 2ln|x + 2| + c1

∴ I1 = x - 2ln|x + 2| + c1

For I2

⇒

Assume x + 1 = t

dt = dx

⇒

Substitute u = √t

dt = 2√t.du

t = u²

⇒

Add and subtract 1 in the above equation:

⇒

⇒

⇒

⇒ 2u - tan ^{- 1}(u) + c2

But u = √t

∴ 2√t - tan ^{- 1}(√t) + c2

Also t = x + 1

∴ 2√(x + 1) - tan ^{- 1}(x + 1) + c2

I = I1 + I2

∴ I = x - 2ln|x + 2| + c1 + 2√(x + 1) - tan ^{- 1}(x + 1) + c2

I = x - 2ln|x + 2| + 2√(x + 1) - tan ^{- 1}(x + 1) + c.

Question 64.

Evaluate the following integrals:

Answer:

Assume

⇒

⇒

Substituting t and dt

⇒

⇒

⇒

⇒

But

⇒

Question 65.

Evaluate the following integrals:

Answer:

Assume x² = t

2x.dx = dt

⇒

Substituting t and dt

⇒

⇒

⇒

⇒

But t = x²

⇒

Question 66.

Evaluate the following integrals:

Answer:

Assume e^x – 1 = t²

d(e^x – 1) = d(t²)

e^x.dx = 2t.dt

⇒

e^x = t² + 1

⇒

Substituting t and dt

⇒

⇒

⇒

⇒

Add and subtract 1 in numerator

⇒

⇒

⇒

⇒

⇒ 2t – 2tan ^{- 1}(t) + c

But t = (e^x – 1)^1/2

⇒ 2(e^x – 1) ^1/2– 2tan ^{- 1}(e^x – 1) ^1/2 + c

Question 67.

Evaluate the following integrals:

Answer:

We can write x² + 2x + 1 + 1 = (x + 1)² + 1

⇒

Assume x + 1 = tant

⇒ dx = sec²t.dx

⇒

⇒ tan²t + 1 = sec²t.

⇒

⇒

⇒ log|sint| + c

⇒

But tant = x + 1

⇒

The final answer is

⇒

Question 68.

Evaluate the following integrals:

Answer:

Assume x³ + 1 = t²

d(x³ + 1) = d(t²)

3x².dx = 2t.dt

⇒

x³ + 1 = t²

⇒

Substituting t and dt

⇒

⇒

⇒

⇒ x³ = t² - 1

⇒

⇒

⇒

⇒

Question 69.

Evaluate the following integrals:

Answer:

Assume 5 - x² = t²

d(5 - x² ) = d(t²)

- 2x.dx = 2t.dt

⇒x dx = - t.dx

⇒

Substituting t and dt

⇒

⇒

⇒ x² = 5 - t²

⇒

⇒

⇒

⇒

Question 70.

Evaluate the following integrals:

Answer:

x = t²

d(x) = 2t.dt

dx = 2t.dt

Substituting t and dt we get

⇒

⇒

⇒

⇒ 2(ln|1 + t|)

But t = √x

⇒ 2( ln|1 + √x|) + c.

Question 71.

Evaluate the following integrals:

Answer:

I =

⇒

Let

⇒

⇒

I =

⇒

⇒

But

⇒

Question 72.

Evaluate the following integrals:

Answer:

Sin⁵x = sin⁴x.sinx

Assume cos x = t

d(cosx) = dt

- sinx.dx = dt

⇒

Substitute t and dt we get

⇒

⇒

⇒

⇒

⇒

⇒ -

⇒

But t = cos x

⇒

---

Exercise 19.10

Question 1.

Evaluate the followign integrals:

Answer:

Substituting, x + 2 = t ⇒dx = dt,

Question 2.

Evaluate the following integrals:

Answer:

Substituting x - 1 = t ⇒ dx = dt,

Question 3.

Evaluate the following integrals:

Answer:

Substituting 3x + 4 = t ⇒ 3dx = dt,

Question 4.

Evaluate the following integrals:

Answer:

Substituting x - 1 = t ⇒dx = dt

Question 5.

Evaluate the following integrals:

Answer:

Substituting x + 2 = t ⇒ dx = dt

Question 6.

Evaluate the following integrals:

Answer:

Substituting x + 1 = t ⇒dx = dt

Question 7.

Evaluate the following integrals:

Answer:

Substituting 1 - x = t ⇒ dx = - dt,

Question 8.

Evaluate the following integrals:

∫ x(1 – x)²³ dx

Answer:

Substituting 1 - x = t ⇒ dx = - dt

Question 9.

Evaluate the following integrals:

Answer:

Question 10.

Evaluate the following integrals:

Answer: