# RS Aggarwal Solutions for Class 12 Chapter 12: Indefinite Integral

We know that the inverse process of differentiation is integration. This exercise covers the concept of integration and important formulas, which are used in solving the problems. The solutions are designed by highly experienced faculty in a stepwise manner as per the exam pattern of the current board. This improves the analytical abilities among students which help them in their exam preparation. In order to score good marks in the board exams, students can access RS Aggarwal Solutions for Class 12 Chapter 12 Indefinite Integral free PDF, from the links which are given below.

## Access RS Aggarwal Solutions for Class 12 Chapter 12: Indefinite Integral

Exercise 12 page: 597

Evaluate:

1.

Solution:

We know that

2.

Solution:

By integration we get

3.

Solution:

4.

Solution:

Solution:

It can be written as

6.

Solution:

It can be written as

Solution:

Solution:

It is given that

By integrating w.r.t x

= – cosec x – tan x + x – sec x + 2 tan x + c

On further calculation

= – cosec x + tan x + x – sec x + c

9. (i) ∫sec x (sec x + tan x) dx

(ii) ∫cosec x (cosec x – cot x) dx

Solution:

(i) ∫sec x (sec x + tan x) dx

It can be written as

= ∫sec2 x + sec x tan x dx

By integrating w.r.t x

= tan x + sec x + c

(ii) ∫cosec x (cosec x – cot x) dx

It can be written as

= ∫cosec2 x – cosec x cot x dx

By integrating w.r.t x

= – cot x + cosec x + c

10.

Solution:

(i) ∫ (tan x + cot x) 2 dx

It can be written as

= ∫ (tan2 x + cot2 x + 2 tan x cot x) dx

So we get

= ∫ ([sec2 x – 1] + [cosec2 x – 1] + 2) dx

On further calculation

= ∫ (sec2 x + cosec2 x) dx

By integrating w.r.t x

= tan x – cot x + c

So we get

= ∫ (sec2 x + 2 tan x sec x) dx

By integrating w.r.t x

= tan x + 2 sec x + c

It can be written as

= ∫ (3 cot x cosec x + 4 cosec2 x) dx

By integrating w.r.t x

= – 3 cosec x – 4 cot x + c

Solution:

So we get

= ∫ (cosec2 x + cosec x cot x) dx

By integrating w.r.t x

= – cot x – cosec x + c

Now multiply and divide by (1 + sin x)

So we get

= ∫ (sec2 x + sec x tan x) dx

By integrating w.r.t x

= tan x + sec x + c

Solution:

It can be written as

= ∫ (tan x sec x – tan2 x) dx

So we get

= ∫ (tan x sec x – [sec2 x – 1]) dx

By integrating w.r.t x

= sec x – tan x + x + c

It can be written as

= ∫ (cosec2 x + cosec x cot x) dx

By integrating w.r.t x

= – cot x – cosec x + c

Solution:

On further calculation

= ∫ (cot x cosec x – cot2 x) dx

It can be written as

= ∫ (cot x cosec x – [cosec2 x – 1]) dx

By integrating w.r.t x

= – cosec x + cot x + x + c

On further calculation

= ∫ (tan x sec x + tan2 x) dx

It can be written as

= ∫ (tan x sec x + [sec2 x – 1]) dx

By integrating w.r.t x

= sec x + tan x – x + c

Solution:

Solution:

Solution:

It is given that

We can write it as

= ∫ √ (sin x + cos x) 2 dx

So we get

= ∫ (sin x + cos x) dx

By integrating w.r.t x

= – cos x + sin x + c

We get

= sin x – cos x + c