Solve ∫ x - a b - x d x .

Let I = ∫ x - a b - x d x Let b - x = t 2 ⇒ - d x = 2 t d t ⇒ d x = - 2 t d t ∴ I = - ∫ b - t 2 - a t 2 · 2 t d t = - 2 ∫ b - a 2 - t 2 d t = - 2 t 2 b - a - t 2 + b - a 2 sin - 1 t b - a + C = - t b - a - t 2 - b - a sin - 1 t b - a + C = - b - x x - a - b - a sin - 1 b - x b - a + C

Solve ∫ x + 1 x + 2 · 1 x + 3 d x .

Let I = ∫ x + 1 x + 2 · 1 x + 3 d x Let x + 1 x + 2 = t 2 ⇒ x + 1 = x t 2 + 2 t 2 ⇒ x = 2 t 2 - 1 1 - t 2 ⇒ d x = ( 1 - t 2 ) · ( 4 t ) - ( 2 t 2 - 1 ) ( - 2 t ) ( 1 - t 2 ) 2 d t = 2 t ( 1 - t 2 ) 2 d t ∴ I = ∫ t 2 · 1 2 t 2 - 1 1 - t 2 + 3 · 2 t 1 - t 2 2 d t = ∫ 2 t 2 2 - t 2 1 - t 2 d t = ∫ 2 1 - t 2 - 4 2 - t 2 d t = 2 · 1 2 . 1 log 1 + t 1 - t - 4 2 . 2 log 2 + t 2 - t + c = l o g 1 + t 1 - t + 2 log 2 + t 2 - t + c where t = x + 1 x + 2

∫ d x 1 + x x - x 2 =

Let I = ∫ d x 1 + x x - x 2 ⇒ I = ∫ d x 1 + x x 1 - x Let x = sin θ ⇒ d x 2 x = cos θ d θ ∴ I = ∫ 2 cos θ d θ 1 + sin θ 1 - sin 2 θ = ∫ 2 d θ 1 + sin θ = 2 ∫ 1 - sin θ 1 - sin 2 θ d θ = 2 ∫ 1 - sin θ cos 2 θ d θ = 2 ∫ sec 2 θ - tan θ sec θ d θ = 2 tan θ - sec θ + c = 2 sin θ - 1 cos θ + c = 2 sin θ - 1 1 - sin 2 θ + c = 2 x - 1 1 - x + c

Let I = ∫ d x 2 + sin x 2 Let A = cos x 2 + sin x ⇒ d A d x = 2 + sin x - sin x - cos x cos x 2 + sin x 2 ⇒ d A d x = - 2 sin x + 1 2 + sin x 2 = - 2 sin x + 2 + 3 2 + sin x 2 = - 2 2 + sin x + 3 2 + sin x 2 Integrating both side with respect to x , we get A = - 2 ∫ d x 2 + sin x + 3 I ⇒ I = A 3 + 2 3 ∫ d x 2 + sin x = 1 3 cos x 2 + sin x + 2 3 ∫ d x 2 + 2 tan x 2 1 + tan 2 x 2 ⇒ I = cos x 3 2 + sin x + 1 3 ∫ s e c 2 x 2 1 + tan 2 x 2 + tan x 2 Let tan x 2 = t ⇒ 1 2 s e c 2 x 2 d x = d t ∴ I = cos x 3 2 + sin x + 2 3 ∫ d t t 2 + t + 1 = cos x 3 2 + sin x + 2 3 ∫ d t t + 1 2 2 + 3 2 2 = cos x 3 2 + sin x + 2 3 · 1 3 / 2 tan - 1 t + 1 / 2 3 / 2 + c = cos x 3 2 + sin x + 4 3 3 tan - 1 2 tan x 2 + 1 3 + c

Let I = ∫ d x x 2 + 1 4 Let x = tan θ ⇒ d x = sec 2 θ d θ ∴ I = ∫ sec 2 θ d θ tan 2 θ + 1 4 = ∫ sec 2 θ d θ sec 8 θ = ∫ cos 6 θ d θ = ∫ cos 3 θ 2 d θ = ∫ cos 3 θ + 3 cos θ 4 2 d θ ∵ cos 3 θ = 4 cos 3 θ - 3 cos θ = 1 16 ∫ cos 2 3 θ + 9 cos 2 θ + 6 cos 3 θ cos θ d θ = 1 16 ∫ 1 + cos 6 θ 2 + 9 1 + cos 2 θ 2 + 3 · cos 4 θ + cos 2 θ d θ = 1 32 ∫ 10 + cos 6 θ + 6 cos 4 θ + 15 cos 2 θ d θ = 1 32 10 θ + sin 6 θ 6 + 6 sin 4 θ 4 + 15 sin 2 θ 2 + C = 1 32 10 θ + 3 sin 2 θ - 4 sin 3 2 θ 6 + 3 2 · 2 sin 2 θ cos 2 θ + 15 2 sin 2 θ + C I = 1 32 10 θ + 8 sin 2 θ - 2 3 sin 3 2 θ + 3 sin 2 θ cos 2 θ + C Now, sin 2 θ = 2 tan θ 1 + tan 2 θ = 2 x 1 + x 2 ∵ x = tan θ And cos 2 θ = 1 - tan 2 θ 1 + tan 2 θ = 1 - x 2 1 + x 2 ∴ I = 1 32 10 tan - 1 x + 8 · 2 x 1 + x 2 - 2 3 2 x 1 + x 2 3 + 3 · 2 x 1 + x 2 1 - x 2 1 + x 2 + C = 5 16 tan - 1 x + x 2 1 + x 2 - x 3 6 1 + x 2 3 + 3 x 16 1 + x 2 2 - 3 16 · x 3 1 + x 2 2 + C , where C is constant of integration.

E M B I B E

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 7 - Indefinite Integrals - II>EXERCISE

K. C. Sinha Solutions for Chapter: Indefinite Integrals - II, Exercise 1: EXERCISE

Author: K. C. Sinha

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha Solutions for Chapter: Indefinite Integrals - II, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 7: Indefinite Integrals - II, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.