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

 

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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 Chapter 13: Methods of Integration

.

Exercise 13a page: 621

Evaluate the following integrals:

Very-Short-Answer Questions

1. ∫ (2x + 9) 5 dx

Solution:

Take 2x + 9 = t

So we get

2 dx = dt

It can be written as

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

2.  ∫ (7 – 3x) 4 dx

Solution:

Take 7 – 3x = t

So we get

– 3 dx = dt

It can be written as

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

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

Solution:

Take 3x – 5 = t

So we get

3 dx = dt

It can be written as

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

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

Solution:

Take 4x + 3 = t

So we get

4 dx = dt

It can be written as

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

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

Solution:

Take 3 – 4x = t

So we get

– 4 dx = dt

It can be written as

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

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

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

Solution:

Take 2x – 3 = t

So we get

2 dx = dt

It can be written as

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

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

Solution:

Take 2x – 1 = t

So we get

2 dx = dt

It can be written as

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

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

Solution:

Take 1 – 3x = t

So we get

– 3 dx = dt

It can be written as

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

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

Solution:

Take 2 – 3x = t

So we get

– 3 dx = dt

It can be written as

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

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

Solution:

Take 3x = t

So we get

3 dx = dt

It can be written as

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

By integrating w.r.t. t

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

Short-Answer Questions

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

Solution:

Take 5 + 6x = t

So we get

6 dx = dt

It can be written as

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

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

Solution:

It is given that

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

By integrating w.r.t. t

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

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

Solution:

Take 2x + 5 = t

So we get

2 dx = dt

By integrating w.r.t. t

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

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

Solution:

Take sin x = t

So we get

cos x dx = dt

By integrating w.r.t. t

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

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

Solution:

Take sin x = t

So we get

cos x dx = dt

By integrating w.r.t. t

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

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

Solution:

Take cos x = t

So we get

– sin x dx = dt

By integrating w.r.t. t

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

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

Solution:

Take sin -1 x = t

So we get

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

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

Solution:

Take tan-1 x = t

So we get

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

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

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

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

Solution:

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

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

By substituting the value of t

= – cot (log x) + c

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

Solution:

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

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

By substituting the value of t

= log (log x) + c

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

Solution:

It is given that

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

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

Solution:

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

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

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

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

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

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

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

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 50

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 51

Solution:

Take ax + b = t

So we get

a dx = dt

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 52

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 53

Solution:

We know that

cos 3x = 4 cos3 x – 3 cos x

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 54

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 55

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 56

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 57

Solution:

Take -1/x = t

So we get

1/x2 dx = dt

By integrating w.r.t. t

∫cos t dt = sin t + c

By substituting the value of t

= – sin 1/x + c

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 58

Solution:

It is given that

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 59

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 60

Solution:

Take e2x – 2 = t

So we get

2 e2x dx = dt

By integrating w.r.t. t

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 61

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 62

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 63

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 64

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

RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13A Image 65

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

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

(ii) ∫cos2 x dx

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

It can be written as

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

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

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

(ii) cot2 (x/2) dx

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

It can be written as

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

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

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

(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

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

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

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

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

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

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

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

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

Solution:

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

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

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

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

Solution:

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

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

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

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 =
RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13B Image 18

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 =
RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13B Image 19

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 =
RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13B Image 20

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 =
RS Aggarwal Solutions for Class 12 Chapter 13 Ex 13B Image 21

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

Solution:

We know that

1 + cos 2x = 2 cos2 x

So it can be written as

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

13. ∫ cos4 x dx

Solution:

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

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

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

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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