Evaluate the following integrals : ∫ d x x ln x

Let ln x = t 1 x d x = d t ⇒ I = ∫ d x x ln x ⇒ I = ∫ d t t ⇒ I = ln t + c ⇒ I = ln ln x + c

Evaluate the following integrals : ∫ x n - 1 x 2 n + a 2 d x

Let I = ∫ x n - 1 d x x 2 n + a 2 Let x n = t ⇒ n x n - 1 d x = d t ∴ I = 1 n ∫ d t t 2 + a 2 ⇒ I = 1 n · 1 a tan - 1 t a + c ⇒ I = 1 a n tan - 1 x n a + c .

Evaluate the following integrals : ∫ ln x + 1 ln x d x x

Let ln x = t Differentiating both sides 1 x d x = d t ⇒ I = ∫ ln x + 1 ln x d x x ⇒ I = ∫ t + 1 t d t ⇒ I = t 2 2 + ln t + c ⇒ ln x 2 2 + ln ln x + c

Find the following integrals : ∫ x 2 1 - x 3 d x

Let 1 - x = t Differentiating both sides - d x = d t I = - ∫ 1 - t 2 t 3 d t Now, integration by parts ⇒ I = - 1 - t 2 ∫ t 1 3 d t - ∫ - 2 1 - t ∫ t 1 3 d t ⇒ I = - 4 3 1 - t 2 t 3 4 - ∫ - 3 2 t - 1 t 4 3 d t ⇒ I = - 4 3 1 - t 2 t 3 4 + 3 2 t 10 3 10 3 - 3 2 t 7 3 7 3 + c ⇒ I = - 4 3 x 2 1 - x 3 4 + 9 20 1 - x 10 3 - 9 14 1 - x 7 3 + c

Find the following integrals : ∫ ln x d x x 1 + ln x

Let 1 + ln x = t 1 x d x = d t ⇒ I = ∫ t - 1 d t t ⇒ I = ∫ t 1 2 d t - ∫ t - 1 2 d t ⇒ I = t 3 2 3 2 - t 1 2 1 2 + c ⇒ I = 2 3 t 3 2 - 2 t 1 2 + c ⇒ I = 2 3 1 + ln x 3 2 - 2 1 + ln x 1 2 + c

Find the following integrals : ∫ cos 5 x sin x d x

I = ∫ cos 5 x sin x d x ⇒ I = ∫ cos x cos 4 x sin x d x ⇒ I = ∫ cos x 1 - sin 2 x 2 sin x d x Let sin x = t ⇒ cos x d x = d t ⇒ I = ∫ 1 - t 2 2 t d t ⇒ I = ∫ 1 + t 4 - 2 t 2 t d t ⇒ I = ∫ t 1 2 + t 9 2 - 2 t 5 2 d t ⇒ I = t 3 2 3 2 + t 11 2 11 2 - 2 t 7 2 7 2 + c ⇒ I = 2 3 t 3 2 + 2 11 t 11 2 + 4 7 t 7 2 + c ⇒ I = 2 3 sin 3 x + 2 11 sin 11 x + 4 7 sin 7 x + c

Find the following integrals : ∫ x 5 1 - x 2 d x

Let 1 - x 2 = t 2 Differentiating both sides ⇒ - 2 x d x = 2 t d t ⇒ x d x = - t d t ⇒ I = ∫ x 5 1 - x 2 d x ⇒ I = ∫ x 4 . x 1 - x 2 d x ⇒ I = ∫ 1 - t 2 2 t d t ⇒ I = ∫ 1 + t 4 - 2 t 2 t d t ⇒ I = ∫ d t t + ∫ t 3 d t - ∫ 2 t d t ⇒ I = ln t + t 4 4 - t 2 + c ⇒ I = ln 1 - x 2 + 1 - x 2 2 4 - 1 - x 2 + c

E M B I B E

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 4 - Indefinite Integrals, Basic Methods of Integration>Integration by Substitution

I A Maron Solutions for Chapter: Indefinite Integrals, Basic Methods of Integration, Exercise 2: Integration by Substitution

Author:I A Maron

I A Maron Mathematics Solutions for Exercise - I A Maron Solutions for Chapter: Indefinite Integrals, Basic Methods of Integration, Exercise 2: Integration by Substitution

Attempt the practice questions on Chapter 4: Indefinite Integrals, Basic Methods of Integration, Exercise 2: Integration by Substitution with hints and solutions to strengthen your understanding. PROBLEMS IN CALCULUS OF ONE VARIABLE solutions are prepared by Experienced Embibe Experts.