Evaluate the following integrals: ∫ d x 5 + 4 cos x

Let I = ∫ d x 5 + 4 cos x Substitute tan x 2 = t ⇒ 1 2 sec 2 x 2 dx = dt ⇒ 1 2 ( 1 + tan 2 x 2 ) dx = dt ⇒ 1 2 ( 1 + t 2 ) dx = dt I = ∫ d x 5 + 4 1 - tan 2 x 2 1 + tan 2 x 2 I = ∫ 1 + tan 2 x 2 5 + 5 tan 2 x 2 + 4 - 4 tan 2 x 2 d x I = ∫ 1 + t 2 t 2 + 9 2 d t 1 + t 2 I = 2 ∫ 1 t 2 + 3 2 d t ∫ d x x 2 + a 2 = 1 a tan - 1 x a I = 2 3 tan - 1 t 3 + c I = 2 3 tan - 1 tan x 2 3 + c Hence, ∫ d x 5 + 4 cos x = 2 3 tan - 1 tan x 2 3 + c

Evaluate ∫ d x 1 + 2 cos x

Let I = ∫ d x 1 + 2 cos x Substitute tan x 2 = t ⇒ 1 2 sec 2 x 2 dx = dt ⇒ 1 2 ( 1 + tan 2 x 2 ) dx = dt ⇒ 1 2 ( 1 + t 2 ) dx = dt I = ∫ d x 1 + 2 1 - tan 2 x 2 1 + tan 2 x 2 I = ∫ 1 + t 2 3 - t 2 2 d t 1 + t 2 I = 2 ∫ 1 3 - t 2 d t ∫ d x a 2 - x 2 = 1 2 a log | a + t a - t | + c I = 2 2 3 log | 3 + tan x 2 3 - tan x 2 | + c I = 1 3 log | 3 + tan x 2 3 - tan x 2 | + c Hence, ∫ d x 1 + 2 cos x = 1 3 log | 3 + tan x 2 3 - tan x 2 | + c

Evaluate ∫ d x 4 cos x - 1

Let I = ∫ d x 4 cos x - 1 I = ∫ d x 4 1 - tan 2 x 2 1 + tan 2 x 2 - 1 = ∫ 1 + tan 2 x 2 3 - 5 tan 2 x 2 d x Let tan x 2 = t sec 2 x 2 d x = 2 d t ( 1 + tan 2 x 2 ) d x = 2 d t ( 1 + t 2 ) d x = 2 d t I = ∫ 2 d t 3 - 5 t 2 = 2 × 1 5 ∫ d t 3 5 - t 2 = 2 5 ∫ d t 3 5 2 - t 2 We know, ∫ d t a 2 - t 2 = 1 2 a log | a + t a - t | + c I = 2 5 × 1 2 3 5 log | 3 5 + t 3 5 - t | + c = 1 15 log | 3 + 5 tan x 2 3 - 5 tan x 2 | + c

Evaluate ∫ d x 4 + 5 sin x

Let I = ∫ d x 4 + 5 sin x I = ∫ d x 4 + 5 2 tan x 2 1 + tan 2 x 2 = ∫ 1 + tan 2 x 2 4 + 4 tan 2 x 2 + 10 tan x 2 d x Let tan x 2 = t sec 2 x 2 d x = 2 d t ( 1 + tan 2 x 2 ) d x = 2 d t ( 1 + t 2 ) d x = 2 d t I = ∫ 2 d t 4 t 2 + 10 t + 4 = ∫ d t 2 t 2 + 5 t + 2 = 1 2 ∫ d t ( t + 5 4 ) 2 - ( 3 4 ) 2 We know, ∫ d t t 2 - a 2 = 1 2 a log | a - t a + t | + c I = 1 2 × 1 2 × 3 4 log | t + 5 4 - 3 4 t + 5 4 + 3 4 | + c I = 1 3 log | t + 1 2 t + 2 | + c I = 1 3 log | tan x 2 + 1 2 tan x 2 + 2 | + c

Evaluate ∫ d x 1 - sin x + cos x

Let I = ∫ d x 1 - sin x + cos x I = ∫ d x 1 - 2 tan x 2 1 + tan 2 x 2 + 1 - tan 2 x 2 1 + tan 2 x 2 = ∫ 1 + tan 2 x 2 2 - 2 tan x 2 d x Let tan x 2 = t sec 2 x 2 d x = 2 d t ( 1 + tan 2 x 2 ) d x = 2 d t ( 1 + t 2 ) d x = 2 d t I = ∫ 2 d t 2 - 2 t = ∫ d t 1 - t We know, ∫ d t a - t = log | 1 - t | - 1 + c I = - log | 1 - tan x 2 | + c

Evaluate ∫ d x 1 - tan x

Let I = ∫ d x 1 - tan x I = ∫ cos x cos x - sin x d x = 1 2 ∫ 2 cos x cos x - sin x d x = 1 2 ∫ cos x + cos x cos x - sin x d x = 1 2 ∫ ( cos x + sin x ) + ( cos x - sin x ) cos x - sin x d x = 1 2 ∫ cos x + sin x d x cos x - sin x + ∫ ( cos x - sin x ) d x cos x - sin x cos x - sin x = t ( - sin x - cos x ) d x = d t ( sin x + cos x ) d x = - d t I = 1 2 ∫ - d t t + ∫ d x I = - 1 2 log | t | + x + c I = - 1 2 log | cos x - sin x | + x + c

Evaluate ∫ d x 1 + cot x

Let I = ∫ d x 1 + cot x Simplifying I = ∫ sin x sin x + cos x d x = 1 2 ∫ 2 sin x sin x + cos x d x = 1 2 ∫ sin x + sin x sin x + cos x d x = 1 2 ∫ ( sin x + cos x ) + ( sin x - cos x ) sin x + cos x d x I = 1 2 ∫ sin x + cos x sin x + cos x d x + 1 2 ∫ sin x + cos x sin x + cos x d x Let sin x + cos x = t ( cos x - sin x ) d x = d t ( sin x - cos x ) d x = - d t = 1 2 ∫ d x + 1 2 ∫ - d t t = 1 2 x - 1 2 log | t | + c I = 1 2 x - 1 2 log | sin x + cos x | + c

E M B I B E

Understanding ISC Mathematics Class 12 Volume 2>Chapter 8 - Integrals>EXERCISE

M L Aggarwal Solutions for Chapter: Integrals, Exercise 9: EXERCISE

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Integrals, Exercise 9: EXERCISE

Attempt the free practice questions on Chapter 8: Integrals, Exercise 9: EXERCISE with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 2 solutions are prepared by Experienced Embibe Experts.