**Chapter 3 Indefinite Integration Ex 3.2(A)**

## Chapter 3 Indefinite Integration Ex 3.2(A)

**I. Integrate the following functions w.r.t. x:**

**Question 1.**

**:**

Solution

Solution

**Question 2.**

**Question 3.Solution:**

**Question 4.Solution:**

**Question 5.**

Solution:

**Question 6.Solution:**

**Question 7.Solution:**

**Question 8.Solution:**

**Question 9. Solution:**

**Question 10.Solution:**

**Question 11. Solution:**

**Question 12.Solution:**

**Question 13.**

Dividing numerator and denominator by cos

^{2}x, we get

**Question 14.Solution:**

**Question 15.Solution:**

**Question 16.Solution:**

**Question 17.Solution:**

**Question 18.Solution:**

**Question 19.Solution:**

**Question 20.Solution:**

**Question 21.Solution:**

**Question 22.Solution:**

**Question 23.Solution:**

**Question 24.Solution:**

**Question 25.Solution:**

**II. Integrate the following functions w.r.t x:**

**Question 1.Solution:**

**Question 2.Solution:**

**Question 3.Solution:**

**Question 4.**

Dividing numerator and denominator of cos

^{2}x, we get

**Question 5.**

= A(3 sin x + 4 cos x) + B (3 cos x – 4 sin x)

∴ sin x + 2 cos x = (3A – 4B) sin x + (4A + 3B) cos x

Equating the coefficients of sin x and cos x on both the sides, we get

3A – 4B = 1 …… (1)

and 4A + 3B = 2 …… (2)

Multiplying equation (1) by 3 and equation (2) by 4, we get

9A – 12B = 3

16A + 12B = 8

On adding, we get

**Question 6.**

= A (2 cos x + 3 sin x) + B (-2 sin x + 3 cos x)

∴ cos x = (2A + 3B) cos x + (3A – 2B) sin x

Equating the coefficients of cos x and sin x on both the sides, we get

2A + 3B = 1 …… (1)

and 3A – 2B = 0 ……. (2)

Multiplying equation (1) by 2 and equation (2) by 3, we get

4A + 6B = 2

9A – 6B = 0

On adding, we get

**Question 7.**

= A(2ex – 5) + B(2ex – 0)

∴ 4ex – 25 = (2A + 2B) ex – 5A

Equating the coefficient of ex and constant on both sides, we get

2A + 2B = 4 …….(1)

and 5A = 25

∴ A = 5

from (1), 2(5) + 2B = 4

∴ 2B = -6

∴ B = -3

**Question 8.Solution:**

**Question 9.**

= A(4e2x – 5) + B(4 . e2x × 2 – 0)

∴ 3e2x + 5 = (4A + 8B) e2x – 5A

Equating the coefficient of e2x and constant on both sides, we get

4A + 8B = 3 …….. (1)

and -5A = 5

∴ A = -1

∴ from (1), 4(-1) + 8B = 3

∴ 8B = 7

∴ B = 7/8

**Question 10.x . cot xSolution:**

**Question 11. Solution:**

Let I = ∫ dx

**Question 12. Solution:**

**Question 13.tan 3x tan 2x tan xSolution:**

Let I = ∫ tan 3x tan 2x tan x dx

tan 3x (1 – tan 2x tan x) = tan 2x + tan x

tan 3x – tan 3x tan 2x tan x = tan 2x + tan x

tan 3x – tan 2x – tan x = tan 3x tan 2x tan x

I = ∫(tan 3x – tan 2x – tan x) dx

= ∫tan3x dx – ∫tan 2x dx – ∫tan x dx

**Question 14. Solution:**

**Question 15.Solution:**

**Question 16.Solution:**

**Question 17.Solution:**