Evaluate the following integral: ∫ 1 cos x + a cos x + b d x

Let I = ∫ 1 cos x + a cos x + b d x = 1 sin a - b ∫ sin a - b cos x + a cos x + b d x = 1 sin a - b ∫ sin x + a - x + b cos x + a cos x + b d x = 1 sin a - b ∫ sin x + a cos x + b - cos x + a sin x + b cos x + a cos x + b ∵ sin x - y = sin x cos y - cos x sin y = 1 sin a - b ∫ tan x + a d x - ∫ tan x + b d x = 1 sin a - b - log cos x + a + log cos x + b + C = 1 sin a - b log cos x + b cos x + a + C

Evaluate the following integral: ∫ 1 x x + 1 d x

Let I = ∫ d x x x + 1 Put x + 1 = u ⇒ 1 2 x d x = d u So, I = ∫ 2 u d u = 2 log u + C = 2 log x + 1 + C

Evaluate the following Integral: ∫ tan 2 x tan 9 x tan 5 x d x

let I = ∫ tan 2 x tan 3 x tan 5 x d x Consider, tan 5 x = tan 2 x + 3 x ⇒ tan 5 x = tan 2 x + tan 3 x 1 - tan 2 x tan 3 x ⇒ tan 5 x 1 - tan 2 x tan 3 x = tan 2 x + tan 3 x = tan 2 x tan 3 x tan 5 x = tan 5 x - tan 2 x - tan 3 x So, I = ∫ tan 5 x - tan 2 x - tan 3 x d x ⇒ I = 1 5 log sec 5 x - 1 2 log sec 2 x - 1 3 log sec 3 x + C

Evaluate the following Integral: ∫ 1 + tan x tan x + θ d x

Let I = 1 + tan x tan x + θ d x Consider, tan θ = tan x + θ - x ⇒ tan θ = tan x + θ - tan x 1 + tan x tan x + θ 1 + tan x tan x + θ = tan x + θ - tan x tan θ So, I = ∫ tan x + θ - tan x tan θ d x = 1 tan θ ∫ tan x + θ d x - ∫ tan x d x = cot θ log sec x + θ - log sec x + C

Evaluate the following integral: ∫ sin 2 x sin x - π 6 sin x + π 6 d x

Let I = ∫ sin 2 x sin x - π 6 sin x + π 6 d x = ∫ sin x + π 6 + x - π 6 sin x - π 6 sin x + π 6 d x = ∫ sin x + π 6 cos x - π 6 + cos x + π 6 sin x - π 6 sin x - π 6 sin x + π 6 d x = ∫ cot x - π 6 d x + ∫ cot x + π 6 d x = - log cosec x - π 6 - log cosec x + π 6 + C

Evaluate the following integral: ∫ e x - 1 + x e - 1 e x + x e d x

let I = ∫ e x - 1 + x e - 1 e x + x e d x = 1 e ∫ e x + e . x e - 1 e x + x e d x Let e x + x e = t ⇒ e x + e . x e - 1 = d t So, I = 1 e ∫ d t t = 1 e log t + C = 1 e log e x + x e + C

Evaluate the following integral: ∫ 1 sin x cos 2 x d x

Let I = ∫ d x sin x cos 2 x = ∫ sin 2 x + cos 2 x sin x cos 2 x d x = ∫ tan x sec x d x + ∫ cosec x d x = sec x + log cosec x - cot x + C

Evaluate the following integral: ∫ 1 cos 3 x - cos x d x

Let I = ∫ 1 cos 3 x - cos x d x = ∫ sin 2 x + cos 2 x - 2 sin 3 x + x 2 sin 3 x - x 2 d x ∵ cos α - cos β = - 2 sin α + β 2 sin α - β 2 = ∫ sin 2 x + cos 2 x - 2 sin 2 x sin x d x I = ∫ sin 2 x + cos 2 x - 2 2 sin x cos x sin x d x = - 1 4 ∫ sin 2 x + cos 2 x cos x sin 2 x d x = - 1 4 ∫ sec x d x + ∫ cot x cosec x d x = - 1 4 log sec x + tan x - cosec x + C

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MATHEMATICS CLASS XII VOLUME-1>Chapter 19 - Indefinite Integrals>EXERCISE

R D Sharma Solutions for Chapter: Indefinite Integrals, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Indefinite Integrals, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 19: Indefinite Integrals, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.