∫ tan x sec 2 x x d x

∫ tan x sec 2 x x d x Let, tan x = t ⇒ sec 2 x · 1 2 x d x = d t ⇒ sec 2 x x d x = 2 d t ∴ I = 2 ∫ t d t = 2 t 2 2 + C = tan 2 x + C , where C is constant of integration.

∫ sec 2 x + 1 x + 1 d x

I = ∫ sec 2 x + 1 x + 1 d x Let, x + 1 = t ⇒ 1 2 x + 1 d x = d t ⇒ d x x + 1 = 2 d t ∴ I = ∫ sec 2 t · 2 d t = 2 tan t + C = 2 tan x + 1 + C , where C is constant of integration.

I = ∫ e x 1 + x sin 2 x e x d x

I = ∫ e x 1 + x sin 2 x e x d x Let, x e x = t ⇒ x e x + e x d x = d t ∴ I = ∫ d t sin 2 t = ∫ cosec 2 t d t = - cot t + C = - cot x e x + C , where C is constant of integration.

∫ cos a x + b a x + b d x

∫ cos a x + b a x + b d x Let a x + b = t ⇒ a 2 a x + b d x = d t ⇒ d x a x + b = 2 a d t ∴ I = ∫ cos t · 2 a d t = 2 a sin t + C = 2 a sin a x + b + C , where C is constant of integration.

E M B I B E

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 1 - Indefinite Integrals - I>EXERCISE

K. C. Sinha Solutions for Chapter: Indefinite Integrals - I, Exercise 7: EXERCISE

Author: K. C. Sinha

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha Solutions for Chapter: Indefinite Integrals - I, Exercise 7: EXERCISE

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