Evaluate: ∫ 0 π / 2 e sin x e sin x + e cos x d x

Let I = ∫ 0 π / 2 e sin x e sin x + e cos x d x . . . i ⇒ I = ∫ 0 π / 2 e sin π 2 - x e sin π 2 - x + e cos π 2 - x d x ⇒ I = ∫ 0 π / 2 e cos x e sin x + e cos x d x . . . ii Adding i & ii , we get 2 I = ∫ 0 π / 2 d x ⇒ 2 I = x 0 π 2 ⇒ I = π 4

Evaluate: ∫ 0 a x x + a - x d x .

Let I = ∫ 0 a x x + a - x d x . . . i ⇒ I = ∫ 0 a a - x a - x + a - a + x d x ⇒ I = ∫ 0 a a - x a - x + x d x . . . . ii Adding i & ii , we get 2 I = ∫ 0 a d x ⇒ 2 I = x 0 a ⇒ I = a 2

Evaluate: ∫ 0 π / 2 a sin x + b cos x sin x + cos x d x

Let I = ∫ 0 π 2 a sin x + b cos x sin x + cos x d x . . . . i ⇒ I = ∫ 0 π 2 asin π 2 - x + bcos π 2 - x sin π 2 - x + cos π 2 - x d x ⇒ I = ∫ 0 π 2 a cos x + b sin x cosx + sin x d x . . . . ii Adding i & ii , we get 2 I = ∫ 0 π 2 b cosx + a sinx + a cos x + b sin x cosx + sin x d x ⇒ 2 I = ∫ 0 π 2 b cosx + a sinx + a cos x + b sin x cosx + sin x d x ⇒ 2 I = ∫ 0 π 2 a + b cosx + sinx cosx + sin x d x ⇒ 2 I = a + b ∫ 0 π 2 d x ⇒ 2 I = a + b x 0 π 2 ⇒ 2 I = a + b π 2 - 0 ⇒ I = a + b π 4

Evaluate: ∫ 0 π 2 sin x - cos x sin x + cos x 2 d x .

Let I = ∫ 0 π 2 sin x - cos x sin x + cos x 2 d x ⇒ I = ∫ 0 π 2 sin π 2 - x - cos π 2 - x sin π 2 - x + cos π 2 - x 2 d x ⇒ I = ∫ 0 π 2 cos x - sin x sin x + cos x 2 d x ⇒ I = - ∫ 0 π 2 sin x - cos x sin x + cos x 2 d x ⇒ I = - I ⇒ I = 0

Evaluate: ∫ 0 π d x 5 + 4 cos 2 x .

Let I = ∫ 0 π d x 5 + 4 cos 2 x We know that cos 2 x = 1 - tan 2 x 1 + tan 2 x ∴ I = ∫ 0 π d x 5 + 4 1 - tan 2 x 1 + tan 2 x ⇒ I = ∫ 0 π d x 5 + 5 tan 2 x + 4 - 4 tan 2 x 1 + tan 2 x ⇒ I = ∫ 0 π sec 2 x d x 9 + tan 2 x Put tan x = t then sec 2 x d x = d t Therefore, I = ∫ 0 0 d t 9 + t 2 ⇒ I = 0

Evaluate: ∫ 0 π 2 2 lnsin x - lnsin 2 x d x

Let I = ∫ 0 π 2 2 lnsin x - lnsin 2 x d x ⇒ I = ∫ 0 π 2 2 lnsin x - ln 2 sin x cos x d x ⇒ I = ∫ 0 π 2 2 lnsin x - ln 2 - ln sin x - ln cos x d x ⇒ I = ∫ 0 π 2 lnsin x - ln 2 - ln cos x d x ⇒ I = ∫ 0 π 2 lnsin x - ln cos x d x - ∫ 0 π 2 ln 2 d x ⇒ I = ∫ 0 π 2 lnsin x - ln cos x d x - πln 2 2 . . . . i Now, I 1 = ∫ 0 π 2 lnsin x - ln cos x d x ⇒ I 1 = ∫ 0 π 2 ln sin π 2 - x - ln cos π 2 - x d x ⇒ I 1 = ∫ 0 π 2 lncos x - lnsin x d x ⇒ I 1 = - ∫ 0 π 2 lnsin x - lncos x d x ⇒ I 1 = - I 1 ⇒ 2 I 1 = 0 ⇒ I 1 = 0 Hence, by i I = - π 2 ln 2

Evaluate: ∫ 0 ∞ ln x + 1 x · d x 1 + x 2 .

Let I = ∫ 0 ∞ ln x + 1 x · d x 1 + x 2 Put x = tan t ⇒ d x = sec 2 t d t I = ∫ 0 π 2 ln tan t + 1 tan t sec 2 t 1 + tan 2 t d t ⇒ I = ∫ 0 π 2 ln tan t + cot t sec 2 t sec 2 t d t ⇒ I = ∫ 0 π 2 ln tan t + cot t d t ⇒ I = ∫ 0 π 2 ln sin t cos t + cos t sin t d t ⇒ I = ∫ 0 π 2 ln sin 2 t + cos 2 t cos t sin t d t ⇒ I = ∫ 0 π 2 ln 2 2 cos t sin t d t ⇒ I = ∫ 0 π 2 ln 2 sin 2 t d t ⇒ I = ∫ 0 π 2 ln 2 d t - ∫ 0 π 2 sin 2 t d t ⇒ I = ln 2 t 0 π 2 - - π 2 ln 2 ⇒ I = πln 2

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 30 - Definite Integration>Exercise-1

Embibe Experts Solutions for Chapter: Definite Integration, Exercise 1: Exercise-1

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Definite Integration, Exercise 1: Exercise-1

Attempt the free practice questions on Chapter 30: Definite Integration, Exercise 1: Exercise-1 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.