Evaluate the following Integral: ∫ e 2 x 1 + e x d x

Let 1 + e x = t ⇒ d 1 + e x = d t ⇒ e x d x = d t I = ∫ e 2 x 1 + e x d x = ∫ e x e x d x 1 + e x = ∫ e x t d t = ∫ t - 1 t d t ∵ 1 + e x = t = ∫ t t - 1 t d t = ∫ 1 - 1 t d t = t - log t + C = 1 + e x - log 1 + e x + C

Evaluate the following Integral: ∫ s e c 2 x x d x

Let x = t ⇒ d x = d t ⇒ 1 2 x d x = d t ⇒ d x x = 2 d t Now, let I = ∫ s e c 2 x x d x = ∫ s e c 2 t 2 d t = 2 ∫ s e c 2 t d t = 2 tan t + C = 2 tan x + C .

Evaluate the following Integral: ∫ tan 3 2 x sec 2 x d x

Use the formula sec 2 x - tan 2 x = 1 I = ∫ tan 3 2 x sec 2 x d x = ∫ tan 2 2 x tan 2 x sec 2 x d x = ∫ sec 2 2 x - 1 tan 2 x sec 2 x d x = ∫ sec 2 2 x tan 2 x sec 2 x d x - ∫ tan 2 x sec 2 x d x I = ∫ sec 2 2 x tan 2 x sec 2 x - sec 2 x 2 + C Let sec 2 x = t ⇒ d sec 2 x = d t ⇒ 2 sec 2 x tan 2 x d x = d t ⇒ sec 2 x tan 2 x d x = d t 2 I = ∫ sec 2 2 x sec 2 x tan 2 x d x - sec 2 x 2 + C = ∫ t 2 d t 2 - sec 2 x 2 + C = 1 2 ∫ t 2 d t - sec 2 x 2 + C = 1 2 t 3 3 - sec 2 x 2 + C = sec 2 x 3 6 - sec 2 x 2 + C

Evaluate the following integral: ∫ x + x + 1 x + 2 d x

Let x + 1 = u 2 ⇒ d x + 1 = u 2 ⇒ d x = 2 u d u I = ∫ x + x + 1 x + 2 d x = ∫ u 2 - 1 + u u 2 - 1 + 2 2 u d u = 2 ∫ u 2 + u - 1 u 2 + 1 u d u = 2 ∫ u 3 + u 2 - u u 2 + 1 d u = 2 ∫ u 3 + u - u + u 2 - u + 1 - 1 u 2 + 1 d u = 2 ∫ u u 2 + 1 - 2 u + u 2 + 1 - 1 u 2 + 1 d u = 2 ∫ u d u - ∫ 2 u u 2 + 1 d u + ∫ 1 d u - ∫ 1 u 2 + 1 d u = 2 ∫ u d u - ∫ d d u u 2 + 1 u 2 + 1 d u + ∫ 1 d u - ∫ 1 u 2 + 1 2 d u Use the formula ∫ f ' x f x d x = log f x and ∫ d x x 2 + a 2 = 1 a tan - 1 x a I = 2 u 2 2 - log u 2 + 1 + u - tan - 1 u + C = 2 x + 1 2 - log x + 1 + 1 + x + 1 - tan - 1 x + 1 + C = 2 x + 1 - log x + 2 + x + 1 - tan - 1 x + 1 + C

Evaluate the following Integral: ∫ 5 5 5 x 5 5 x 5 x d x

Let 5 5 5 x = t ⇒ d 5 5 5 x = d t ⇒ 5 5 5 x 5 5 x 5 x log 5 3 d x = d t ⇒ 5 5 5 x 5 5 x 5 x d x = d t log 5 3 Now, let I = ∫ 5 5 5 x 5 5 x 5 x d x = ∫ d t log 5 3 = 1 log 5 3 ∫ 1 d t = 1 log 5 3 t + C = 5 5 5 x log 5 3 + C .

Evaluate the following Integral: ∫ 1 x x 4 - 1

Let x 2 = u ⇒ d x 2 = d u ⇒ 2 x d x = d u ⇒ d x = d u 2 x Now, let I = ∫ 1 x x 4 - 1 = ∫ 1 x x 2 2 - 1 d u 2 x = 1 2 ∫ 1 x 2 x 2 2 - 1 d u = 1 2 ∫ 1 u u 2 - 1 d u = 1 2 sec - 1 u + C = 1 2 sec - 1 x 2 + C

Evaluate the following Integral: ∫ e x - 1 d x

Let e x - 1 = t ⇒ d e x - 1 = t 2 ⇒ e x d x = 2 t d t ⇒ d x = 2 t d t e x ⇒ d x = 2 t t 2 + 1 d t Now, let I = ∫ e x - 1 d x = ∫ t 2 2 t t 2 + 1 d t = 2 ∫ t 2 t 2 + 1 d t Add and subtract 1 in numerator = 2 ∫ t 2 + 1 - 1 t 2 + 1 d t = 2 ∫ t 2 + 1 t 2 + 1 - 1 t 2 + 1 d t = 2 ∫ 1 - 1 t 2 + 1 d t = 2 t - tan - 1 t + C = 2 e x - 1 - tan - 1 e x - 1 + C .

Evaluate the following Integral: ∫ sin 5 x cos 4 x d x

Let cos x = t d cos x = d t - sin x d x = d t d x = d t - sin x Now, let I = ∫ sin 5 x cos 4 x d x = ∫ sin 5 x cos 4 x × d t - sin x = - ∫ sin 4 x cos 4 x d t = - ∫ 1 - t 2 2 t 4 d t Use the formula a - b 2 = a 2 - 2 a b + b 2 = - ∫ 1 - 2 t 2 + t 4 t 4 dt = - ∫ 1 t 4 - 2 t 2 t 4 + t 4 t 4 dt = - ∫ 1 t 4 - 2 t 2 + 1 dt = - ∫ t - 4 + 1 - 4 + 1 - 2 t - 2 + 1 - 2 + 1 + t + C = - ∫ t - 3 - 3 - 2 t - 1 - 1 + t + C = t - 3 3 - 2 t - t + C = 1 3 cos 3 x - 2 cos x - cos x + C

E M B I B E

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.