RS Aggarwal Solutions for Class 12 Chapter 13: Methods of Integration Exercise 13B

In this exercise, the problems are based on the concept of integration using trigonometric identities. Students can solve the exercise wise problems more easily using the solutions PDF as a reference guide. Using the solutions while solving problems improves the analytical thinking and problem solving abilities among students. Students can boost their confidence in solving the exercise wise problems by using RS Aggarwal Solutions Class 12 Maths Chapter 13 Methods of Integration Exercise 13B PDF from the links, which are available here.

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

 

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Access other exercise solutions of Class 12 Maths Chapter 13: Methods of Integration

Exercise 13A Solutions

Exercise 13C Solutions

Access RS Aggarwal Solutions for Class 12 Chapter 13: Methods of Integration Exercise 13B

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

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