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

The problems of exercise 13C are based on the concept of integration by parts. The theorem which is explained under this exercise mainly provides an exact idea about the steps to be followed in solving problems. Students can clear their doubts instantly by solving problems using solutions prepared by our experts. It improves the time management skills among students, which are important from the exam point of view. RS Aggarwal Solutions Class 12 Maths Chapter 13 Methods of Integration Exercise 13C PDF links are available below.

## RS Aggarwal Solutions for Class 12 Maths Chapter 13: Methods of Integration Exercise 13C Download PDF                       ### Access other exercise solutions of Class 12 Maths Chapter 13: Methods of Integration

Exercise 13A Solutions

Exercise 13B Solutions

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

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