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

Integration by Substitution is the main concept which is covered under Exercise 13A of Chapter 13. The problems are solved by teachers having a high knowledge of concepts with the aim of helping students score well in the board exam. The solved examples before each exercise give an idea to the students about the kind of questions that would appear in the exam. Students who aspire to perform well in the board exam can download RS Aggarwal Solutions Class 12 Maths Chapter 13 Methods of Integration Exercise 13A PDF using links, which are provided below.

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

 

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

Exercise 13B Solutions

Exercise 13C Solutions

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

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

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

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