# RS Aggarwal Solutions for Class 12 Maths Chapter 13: Methods of Integration

The various methods of integrating the given set of questions are explained in brief under this chapter. Students who have difficulty in solving these problems can use the solutions PDF to clarify their doubts. It mainly improves problem-solving abilities among students in order to boost their exam preparation. The solutions are prepared by experts in a step by step manner based on CBSE exam pattern. For a better hold on the concepts covered under this chapter, RS Aggarwal Solutions for Class 12 Chapter 13 Methods of Integration PDF links are available below.

## RS Aggarwal Solutions for Class 12 Maths Chapter 13: Methods of Integration Download PDF

### Exercises of RS Aggarwal Solutions Class 12 Maths Chapter 13 – Methods of Integration

Exercise 13A Solutions

Exercise 13B Solutions

Exercise 13C Solutions

## Access RS Aggarwal Solutions for Class 12 Maths Chapter 13: Methods of Integration

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Exercise 13a page: 621

Evaluate the following integrals:

1. ∫ (2x + 9) 5 dx

Solution:

Take 2x + 9 = t

So we get

2 dx = dt

It can be written as

2.  ∫ (7 – 3x) 4 dx

Solution:

Take 7 – 3x = t

So we get

– 3 dx = dt

It can be written as

Solution:

Take 3x – 5 = t

So we get

3 dx = dt

It can be written as

Solution:

Take 4x + 3 = t

So we get

4 dx = dt

It can be written as

Solution:

Take 3 – 4x = t

So we get

– 4 dx = dt

It can be written as

Solution:

Take 2x – 3 = t

So we get

2 dx = dt

It can be written as

Solution:

Take 2x – 1 = t

So we get

2 dx = dt

It can be written as

Solution:

Take 1 – 3x = t

So we get

– 3 dx = dt

It can be written as

Solution:

Take 2 – 3x = t

So we get

– 3 dx = dt

It can be written as

Solution:

Take 3x = t

So we get

3 dx = dt

It can be written as

By integrating w.r.t. t

Solution:

Take 5 + 6x = t

So we get

6 dx = dt

It can be written as

Solution:

It is given that

By integrating w.r.t. t

Solution:

Take 2x + 5 = t

So we get

2 dx = dt

By integrating w.r.t. t

Solution:

Take sin x = t

So we get

cos x dx = dt

By integrating w.r.t. t

Solution:

Take sin x = t

So we get

cos x dx = dt

By integrating w.r.t. t

Solution:

Take cos x = t

So we get

– sin x dx = dt

By integrating w.r.t. t

Solution:

Take sin -1 x = t

So we get

Solution:

Take tan-1 x = t

So we get

Solution:

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

∫cos t dt = sin t + c

By substituting the value of t

= sin (log x) + c

Solution:

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

By substituting the value of t

= – cot (log x) + c

Solution:

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

By substituting the value of t

= log (log x) + c

Solution:

It is given that

Solution:

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

Solution:

Take √x = t

So we get

1/2√x dx = dt

By integrating w.r.t. t

∫cos t 2 dt = 2 sin t + c

By substituting the value of t

= 2 sin √x + c

Solution:

Take tan x = t

So we get

sec2 x dx = dt

By integrating w.r.t. t

∫et dt = et + c

By substituting the value of t

= e tan x + c

Solution:

Take cos2 x = t

So we get

– sin 2x dx = dt

By integrating w.r.t. t

∫- et dt = – et + c

By substituting the value of t

Solution:

Take ax + b = t

So we get

a dx = dt

It can be written as

Solution:

We know that

cos 3x = 4 cos3 x – 3 cos x

It can be written as

Solution:

Take -1/x = t

So we get

1/x2 dx = dt

By integrating w.r.t. t

∫et dt = et + c

By substituting the value of t

Solution:

Take -1/x = t

So we get

1/x2 dx = dt

By integrating w.r.t. t

∫cos (-t) dt = ∫cos t dt=  sin t + c     [Since, cos is an even function]

By substituting the value of t

= – sin 1/x + c

Solution:

It is given that

Solution:

Take e2x – 2 = t

So we get

2 e2x dx = dt

By integrating w.r.t. t

Solution:

Take log (sin x) = t

So we get

cos x/ sin x dx = dt

By cross multiplication

cot x dx = dt

By integrating w.r.t. t

Solution:

Take log (sin x) = t

So we get

cos x/ sin x dx = dt

By cross multiplication

cot x dx = dt

By integrating w.r.t. t

∫1/t dt = log t + c

By substituting the value of t

= log (log sin x) + c

Solution:

Take x2 + 1 = t

So we get

2x dx = dt

By integrating w.r.t. t

∫sin t dt = – cos t + c

By substituting the value of t

= – cos (x2 + 1) + c

Exercise 13B page: 658

Evaluate the following integrals:

1. (i) ∫sin2 x dx

(ii) ∫cos2 x dx

Solution:

(i) ∫sin2 x dx

Here, we know that 1 – cos 2x = 2 sin2 x

It can be written as

(ii) ∫cos2 x dx

Here, we know that 1 + cos 2x = 2 cos2 x

It can be written as

2. (i) cos2 (x/2) dx

(ii) cot2 (x/2) dx

Solution:

(i) cos2 (x/2) dx

Here, we know that 1 + cos x = 2 cos2 (x/2)

It can be written as

(ii) cot2 (x/2) dx

Here, we know that cosec2 x – cot2 x = 1

It can be written as

3. (i) ∫ sin2 nx dx

(ii) ∫ sin5 x dx

Solution:

(i) ∫ sin2 nx dx

Here, we know that 1 – cos 2nx = 2 sin2 nx

It can be written as

(ii) ∫ sin5 x dx

Here, we know that 1 – cos2 x = sin2 x

It can be written as

∫sin5 x dx = ∫(1 – cos2 x)2 sin x dx

Take cos x = t

So we get

– sin x dx = dt

By considering cos x = t we get

∫(1 – cos2 x)2 sin x dx = – ∫(1 – t2)2 dt

Here

– ∫(1 – t2)2 dt = – ∫(1 + t4 – 2t2) dt

We get

– ∫(1 – t2)2 dt = – ∫ dt + ∫ 2t2 dt – ∫ t4 dt

By integrating w.r.t t

4. ∫ cos3 (3x + 5) dx

Solution:

It is given that

∫ cos3 (3x + 5) dx

By substituting 3x + 5 = u

We get 3 dx = du

Here dx = du/3

So we get

Now by re-substituting the value of t = sin u and u = 3x + 5

5. ∫ sin7 (3 – 2x) dx

Solution:

We can write it as

= – ∫ sin7 (2x – 3) dx

By substituting the value of 2x – 3 = u

So we get

2 dx = du where dx = du/2

Solution:

Solution:

8. ∫ sin 3x cos 4x dx

Solution:

It is given that

∫ sin 3x cos 4x dx

Using the formula sin x × cos y=1/2(sin(x + y)-sin(y-x))

∫ sin 3x cos 4x dx = ½ ∫ (sin 7x – sin x) dx

On further simplification

∫ sin 3x cos 4x dx = ½ ∫ sin 7x dx – ½ ∫ sin x dx

By integrating w.r.t. x

∫ sin 3x cos 4x dx =

9. ∫ cos 4x cos 3x dx

Solution:

It is given that

∫ cos 4x cos 3x dx

Using the formula cos x × cos y=1/2(cos(x + y)+cos(x-y))

∫ cos 4x cos 3x dx = ½ ∫ (cos 7x + cos x) dx

On further simplification

∫ cos 4x cos 3x dx = ½ ∫ cos 7x dx + ½ ∫ cos x dx

By integrating w.r.t. x

∫ cos 4x cos 3x dx =

10. ∫ sin 4x sin 8x dx

Solution:

It is given that

∫ sin 4x sin 8x dx

Using the formula sin x × sin y=1/2(cos(y-x)-cos(y + x))

∫ sin 4x sin 8x dx = ½ ∫ (cos 4x – cos 12x) dx

On further simplification

∫ sin 4x sin 8x dx = ½ ∫ cos 4x dx – ½ ∫ cos 12x dx

By integrating w.r.t. x

∫ sin 4x sin 8x dx =

11. ∫ sin 6x cos x dx

Solution:

It is given that

∫ sin 6x cos x dx

Using the formula sin x × cos y=1/2(sin(y + x)-sin(y-x))

∫ sin 6x cos x dx = ½ ∫ (sin 7x – sin (-5x)) dx

On further simplification

∫ sin 6x cos x dx = ½ ∫ sin 7x dx + ½ ∫ sin 5x dx

By integrating w.r.t. x

∫ sin 6x cos x dx =

Solution:

We know that

1 + cos 2x = 2 cos2 x

So it can be written as

13. ∫ cos4 x dx

Solution:

14. ∫ cos 2x cos 4x cos 6x dx

Solution:

It is given that

∫ cos 2x cos 4x cos 6x dx

By multiplying and dividing by 2

= ½ ∫ 2 cos 2x. cos 4x. cos 6x dx

We can write it as

= ½ ∫ cos 2x (2 cos 4x cos 6x) dx

On further calculation

= ½ ∫ cos 2x [cos (4x + 6x) + cos (4x – 6x)] dx

So we get

= ½ ∫ cos 2x (cos 10x + cos 2x) dx

By further simplification

= ½ ∫ cos 2x cos 10 x dx + ½ ∫ cos2 2x dx

It can be written as

= ½. ½ ∫ 2 cos 2x. cos 10x dx + ½ ∫ (1 + cos 4x)/2 dx

By further multiplication

= ¼ ∫ [cos (2x + 10x) + cos (2x – 10x)] dx + ¼ ∫ (1 + cos 4x) dx

So we get

= ¼ ∫ [cos 12x + cos 8x] dx + ¼ [x + sin 4x/4]

By integrating w.r.t x

= ¼ [sin 12x/12 + sin 8x/8] + ¼x + sin 4x/16 + c

We get

= sin 12x/48 + sin 8x/32 + x/4 + sin 4x/16 + c

15. ∫ sin3 x cos x dx

Solution:

It is given that

∫ sin3 x cos x dx

By taking sin x = t

We get

cos x = dt/dx

It can be written as

cos x dx = dt

So we get

= ∫ t3 dt

By integrating w.r.t t

= t4/4 + c

By substituting the value of t

= sin4 x/4 + c

16. ∫ sec4 x dx

Solution:

It is given that

∫ sec4 x dx

We can write it as

= ∫ sec2 x sec2 x dx

So we get

= ∫ (1 + tan2 x) sec2 x dx

Take tan x = t

By differentiation we get

sec2 x dx = dt

It can be written as

= ∫ (1 + t2) dt

By integrating w.r.t. t

= t + t3/3 + c

By substituting the value of t

= tan x + tan3 x/3 + c

17. ∫ cos3 x sin4x dx

Solution:

It is given that

∫ cos3 x sin4x dx

We can write it as

= ∫ sin4 x cos2 x cos x dx

By taking sin4 x as common

= ∫ sin4 x (1 – sin2 x) cos x dx

Take sin x = t

We get cos x = dx/dt

Here cos x dx = dt

It can be written as

= ∫ t4 (1 – t2) dt

On further simplification

= ∫ t4 dt – ∫ t6 dt

By integration w.r.t. t

= t5/5 – t7/7 + c

By substituting the value of t

= sin5 x/5 – sin7 x/7 + c

Exercise 13C page: 678

Evaluate the following integrals:

1. ∫ x ex dx

Solution:

It is given that

∫ x ex dx

We can write it as

By integrating w.r.t x

= x ex – ∫ 1 ex dx

So we get

= x ex – ex + c

By taking ex as common

= ex (x – 1) + c

2. ∫ x cos x dx

Solution:

It is given that

∫ x cos x dx

We can write it as

By integrating w.r.t x

= x sin x – ∫ 1 sin x dx

So we get

= x sin x + cos x + c

3. ∫ x e2x dx

Solution:

It is given that

∫ x e2x dx

We can write it as

4. ∫ x sin 3x dx

Solution:

It is given that

∫ x sin 3x dx

We can write it as

5. ∫ x cos 2x dx

Solution:

It is given that

∫ x cos 2x dx

We can write it as

So we get

6. ∫ x log 2x dx

Solution:

It is given that

∫ x log 2x dx

We can write it as

7. ∫ x cosec2 x dx

Solution:

It is given that

∫ x cosec2 x dx

We can write it as

By integrating w.r.t x

= x (- cot x) – ∫ 1 (- cot x) dx

So we get

= – x cot x + ∫ cot x dx

Again by integration we get

= – x cot x + log |sin x| + c

8. ∫ x2 cos x dx

Solution:

It is given that

∫ x2 cos x dx

We can write it as

By integrating w.r.t x

= x2 sin x – ∫ 2x sin x dx

So we get

= x2 sin x – 2 [∫ x sin x dx]

Now apply by the part method

= x2 sin x – 2 ∫ x sin x dx

Again by integration

We get

= x2 sin x – 2 [x (- cos x) – ∫ 1. (- cos x) dx]

By integrating w.r.t x

= x2 sin x – 2 (- x cos x + sin x) + c

On further simplification

= x2 sin x + 2x cos x – 2 sin x + c

9. ∫ x sin2 x dx

Solution:

It is given that

∫ x sin2 x dx

We can write it as

sin2 x = (1 + cos 2x)/ 2

By integration

It can be written as

10. ∫ x tan2 x dx

Solution:

It is given that

∫ x tan2 x dx

We can write it as

tan2 x = sec2 x – 1

So we get

∫ x tan2 x dx = ∫ x ( sec2 x – 1) dx

By further simplification

= ∫ x sec2 x dx – ∫ x dx

Taking first function as x and second function as sec2 x

∫ x sec2 x dx – ∫ x dx

We get

By integration w.r.t x

= {x tan x – ∫ tan x dx} – x2/ 2

It can be written as

= x tan x – log |sec x| – x2/ 2 + c

So we get

= x tan x – log |1/cos x| – x2/ 2 + c

Here

= x tan x + log |cos x| – x2/ 2 + c

11. ∫ x2 ex dx

Solution:

It is given that

∫ x2 ex dx

By integrating w.r.t x

We get

= x2 ex – 2 [x ex – ∫1. ex dx]

By further simplification

= x2 ex – 2 [x ex – ex] + c

By multiplication

= x2 ex – 2x ex + 2 ex + c

By taking ex as common

= ex (x2 – 2x + 2) + c

13. ∫ x2 e3x dx

Solution:

It is given that

∫ x2 e3x dx

We can write it as

14. ∫ x2 sin2 x dx

Solution:

It is given that

∫ x2 sin2 x dx

We know that

Now by integrating the second part we get

15. ∫ x3 log 2x dx

Solution:

It is given that

∫ x3 log 2x dx

We can write it as

16. ∫ x log (x + 1) dx

Solution:

It is given that

Solution:

It is given that

Solution:

It is given that

We can write it as

= t et – ∫1. et dt

So we get

= t et – et + C

Now by replacing t with x2

19. ∫ x sin3 x dx

Solution:

It is given that

∫ x sin3 x dx

We know that

sin 3x = 3 sin x – 4 sin3 x

It can be written as

sin3 x = (3 sin x – sin 3x)/4

We get

So we get

20. ∫ x cos3 x dx

Solution:

It is given that

∫ x cos3 x dx

We know that

cos3 x = (cos 3x + 3 cos x)/4

It can be written as

21. ∫ x3 cos x2 dx

Solution:

It is given that

∫ x3 cos x2 dx

We can write it as

∫ x x2 cos x2 dx

Take x2 = t

So we get 2x dx = dt

x dx = dt/2

22. ∫ sin x log (cos x) dx

Solution:

It is given that

∫ sin x log (cos x) dx

Consider first function as log (cos x) and second function as sin x

So we get

= – cos x log (cos x) – ∫ sin x dx

Again by integrating the second term

= – cos x log (cos x) + cos x + c

23. ∫ x sin x cos x dx

Solution:

It is given that

∫ x sin x cos x dx

We know that

sin 2x = 2 sin x cos x

It can be written as

Consider first function as x and second function as sin 2x

24. ∫ cos √x dx

Solution:

It is given that

∫ cos √x dx

Take √x = t

So we get

1/2√x dx = dt

By cross multiplication

dx = 2 √x dt

Here dx = 2t dt

It can be written as

∫ cos √x dx = 2 ∫t cos t dt

Take first function as t and second function as cos t

By integrating w.r.t t

= 2 (t sin t – ∫ sin t dt)

We get

= 2t sin t + 2 cos t + c

Now by substituting t as √x

= 2√x sin √x + 2 cos √x + c

Taking 2 as common

= 2 (cos √x + √x sin √x) + c

25. ∫ cosec3 x dx

Solution:

It is given that

∫ cosec3 x dx

We can write it as

∫ cosec3 x dx = ∫ cosec x cosec2 x dx

So we get

By integration we get

= cosec x (- cot x) – ∫ (- cosec x cot x) (- cot x) dx

It can be written as

= – cosec x cot x – ∫ cosec x cot2 x dx

Here cot2 x = cosec2 x – 1

We get

= – cosec x cot x – ∫ cosec x (cosec2 x – 1) dx

By further simplification

= – cosec x cot x – ∫ cosec3 x dx + ∫ cosec x dx

We know that ∫ cosec3 x dx = 1

∫cosec3 x dx = – cosec x cot x – ∫ cosec3 x dx + ∫ cosec x dx

On further calculation

2 ∫cosec3 x dx = – cosec x cot x + ∫ cosec x dx

By integration we get

2 ∫cosec3 x dx = – cosec x cot x + log |tan x/2| + c

Here

∫cosec3 x dx = – ½ cosec x cot x + ½ log |tan x/2| + c

27. ∫ sin x log (cos x) dx

Solution:

It is given that

∫ sin x log (cos x) dx

Take cos x = t

So we get

– sin x dx = dt

It can be written as

∫ sin x log (cos x) dx = – ∫log t dt = – ∫1. log t dt

Consider first function as log t and second function as 1

By integrating w.r.t t

= – log t. t + ∫1/t. t dt

Again by integrating the second term

= – t log t + t + c

Now replace t as cos x

= t (- log t + 1) + c

We get

= cos x (1 – log (cos x)) + c

Solution:

Take log x = t

So we get

1/x dx = dt

Here

Again by integrating the second term we get

= t log t – t + c

Replace t with log x

= log x. log (log x) – log x + c

By taking log x as common

= log x (log (log x) – 1) + c

29. ∫ log (2 + x2) dx

Solution:

It is given that ∫ log (2 + x2) dx

We can write it as

= ∫1. log (2 + x2) dx

Consider first function as log (2 + x2) and second function as 1

So we get

Solution:

Consider log x = t

Where x = et

So we get

dx = et dt

We can write it as

32. ∫ e –x cos 2x cos 4x dx

Solution:

Using the formula

cos A cos B = ½ [cos (A + B) + cos (A – B)]

We get

cos 4x cos 2x = ½ [cos (4x + 2x) + cos (4x – 2x)] = ½ [cos 6x + cos 2x]

By applying in the original equation

∫ e –x cos 2x cos 4x dx = ∫ e –x (½ [cos 6x + cos 2x])

= ½ [∫e –x cos 6x dx + ∫e –x cos 2x dx]

Taking first function as cos 6x and cos 2x and second function as e –x

Now by solving both parts separately

33. ∫e √x dx

Solution:

Consider √x = t

We get

It can be written as

dx = 2 √x dt

where dx = 2t dt

We can replace it in the equation

∫e √x dx = ∫et 2t dt

So we get

= 2 ∫t et dt

Consider the first function as t and second function as et

By integration w.r.t. t

= 2 (t et – ∫1. et dt)

We get

= 2 (t et – et) + c

Taking et as common

= 2 et (t – 1) + c

Substituting the value of t we get

= 2 e √x (√x – 1) + c

34. ∫ e sin x sin 2x dx

Solution:

We know that

sin 2x = 2 sin x cos x

So we get

∫ e sin x sin 2x dx = 2 ∫ e sin x sin x cos x dx

Take sin x = t

So we get cos x dx = dt

It can be written as

2 ∫ e sin x sin x cos x dx = 2 ∫ e t t dt

Consider first function as t and second function as et

By integrating w.r.t. t

= 2 (t et – ∫1. et dt)

We get

= 2 (t et – et) + c

Here

= 2 et (t – 1) + c

By substituting the value of t

= 2 esin x (sin x – 1) + C

Solution:

Take sin -1 x = t

Here x = sin t

So we get

By integration we get

= t (- cos t) – ∫1. (- cos t) dt

We get

= – t cos t + sin t + c

It can be written as

cos t = √1 – sin2 t

Here we get

= – t (√1 – sin2 t) + sin t + c

By replacing t as sin-1 x

= – sin-1 x (√1 – x2) + x + c