RS Aggarwal Solutions for Class 12 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 Chapter 13: Methods of Integration Exercise 13C Download PDF

 

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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 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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 1

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 2

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 3

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 4

4. ∫ x sin 3x dx

Solution:

It is given that

∫ x sin 3x dx

We can write it as

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 5

5. ∫ x cos 2x dx

Solution:

It is given that

∫ x cos 2x dx

We can write it as

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 6

So we get

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 7

6. ∫ x log 2x dx

Solution:

It is given that

∫ x log 2x dx

We can write it as

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 8

7. ∫ x cosec2 x dx

Solution:

It is given that

∫ x cosec2 x dx

We can write it as

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 9

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 10

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 11

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 12

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 13

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 14

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 15

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 16

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 17

14. ∫ x2 sin2 x dx

Solution:

It is given that

∫ x2 sin2 x dx

We know that

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 18

Now by integrating the second part we get

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 19

15. ∫ x3 log 2x dx

Solution:

It is given that

∫ x3 log 2x dx

We can write it as

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 20

16. ∫ x log (x + 1) dx

Solution:

It is given that

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 21

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 22

Solution:

It is given that

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 23

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 24

Solution:

It is given that

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 25

We can write it as

= t et – ∫1. et dt

So we get

= t et – et + C

Now by replacing t with x2

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 26

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 27

So we get

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 28

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 29

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 30

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 31

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 32

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 33

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 34

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 35

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 36

By integrating w.r.t t

= – log t. t + ∫1/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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 37

Solution:

Take log x = t

So we get

1/x dx = dt

Here

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 38

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 39

So we get

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 40

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 41

Solution:

Consider log x = t

Where x = et

So we get

dx = et dt

We can write it as

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 42

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 43

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 44

33. ∫e √x dx

Solution:

Consider √x = t

We get

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 45

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 46

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 47

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 48

Solution:

Take sin -1 x = t

Here x = sin t

So we get

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13C Image 49

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

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