Evaluate ∫ cos 5 x sin x d x

Let sin x = t then cos x d x = d t We know that, sin 2 y + cos 2 y = 1 Let the given integral be I = ∫ cos 5 x sin x d x = ∫ 1 - t 2 2 t d t = ∫ 1 + t 4 - 2 t 2 t d t = ∫ 1 t d t + ∫ t 3 d t - ∫ 2 t d t I = - 1 t 2 + t 4 4 - t 2 Now putting t = sin x in the equation I = - 1 sin 2 x + sin 4 x 4 - sin 2 x

Evaluate ∫ 0 π 2 cos x ( 1 + sin x ) ( 2 + sin x ) d x

Let, I = ∫ 0 π 2 cos x ( 1 + sin x ) ( 2 + sin x ) d x Let sin x = t Differentiating both the sides, we get cos x d x = d t When x = 0 , then t = 0 When x = π 2 , then t = 1 ∴ I = ∫ 0 1 1 1 + t 2 + t d t = ∫ 0 1 1 1 + t - 1 2 + t d t ⇒ I = log 1 + t - log 2 + t 0 1 ⇒ I = log 1 + t 2 + t 0 1 ⇒ I = log 2 3 - log 1 2 ⇒ I = log 2 3 1 2 ∴ I = log 4 3 .

Evaluate ∫ π 4 3 π 4 x 1 + sin x d x

Let the given integral is I = ∫ π 4 3 π 4 x 1 + sin x d x . .............(i) We know that, ∫ a b f x d x = ∫ a b f a + b - x d x ⇒ I = ∫ π 4 3 π 4 π - x 1 + sin π - x d x ⇒ I = ∫ π 4 3 π 4 π - x 1 + sin x d x . ..........(ii) Adding (i) and (ii) we get 2 I = ∫ π 4 3 π 4 π 1 + sin x d x We know that, sin 2 θ + cos 2 θ = 1 ⇒ 2 I = ∫ π 4 3 π 4 π sin 2 x 2 + cos 2 x 2 + 2 sin x 2 cos x 2 d x ⇒ 2 I = ∫ π 4 3 π 4 π sin x 2 + cos x 2 2 d x ⇒ 2 I = ∫ π 4 3 π 4 π sec 2 x 2 1 + tan x 2 2 d x Let 1 + tan x 2 = t ⇒ d t = 1 2 s e c 2 x 2 d x ⇒ 2 I = 2 π ∫ π 4 3 π 4 d t t 2 ⇒ 2 I = 2 π - 1 t x = π 4 x = 3 π 4 ⇒ 2 I = 2 π - 1 1 + tan x 2 π 4 3 π 4 ⇒ I = π 1 1 + tan π 8 - 1 1 + tan 3 π 8 ⇒ I = π ( 2 - 1 )

Evaluate ∫ log x 2 x d x .

Let I = ∫ log x 2 x d x Put log x = t Differentiate both the sides with respect to x , we get 1 x d x = d t . . . 3 Now, we have I = ∫ t 2 d t ⇒ I = t 3 3 + C ⇒ I = log x 3 3 + C , C integration constant.

Evaluate ∫ 0 π 4 log ( 1 + tan x ) d x .

Let I = ∫ 0 π 4 log 1 + tan x d x ⇒ I = ∫ 0 π 4 log 1 + tan π 4 - x d x ⇒ I = ∫ 0 π 4 log 1 + 1 - tan x 1 + tan x d x ⇒ I = ∫ 0 π 4 log 2 1 + tan x d x ⇒ I = ∫ 0 π 4 log 2 - log 1 + tan x d x ⇒ I = ∫ 0 π 4 log 2 d x - ∫ 0 π 4 log 1 + tan x d x ⇒ I = log 2 ∫ 0 π 4 d x - I ⇒ 2 I = log 2 x 0 π 4 ⇒ 2 I = π 4 log 2 ⇒ I = π 8 log 2

Evaluate ∫ x + 2 x 2 + 2 x + 3 d x .

We have to evaluate ∫ x + 2 x 2 + 2 x + 3 d x Let I = ∫ x + 1 + 1 x 2 + 2 x + 3 d x = 1 2 ∫ 2 x + 1 x 2 + 2 x + 3 d x + ∫ 1 x 2 + 2 x + 3 d x Let I 1 = 1 2 ∫ 2 x + 1 x 2 + 2 x + 3 d x , I 2 = ∫ 1 x 2 + 2 x + 3 d x , so I = I 1 + I 2 Consider, I 1 = 1 2 ∫ 2 x + 1 x 2 + 2 x + 3 d x Put x 2 + 2 x + 3 = t ⇒ 2 x + 2 d x = d t I 1 = 1 2 ∫ 1 t d t = 1 2 t - 1 2 + 1 - 1 2 + 1 + C 1 = t + C 1 ⇒ I 1 = x 2 + 2 x + 3 + C 1 Now, I 2 = ∫ 1 x 2 + 2 x + 3 d x = ∫ 1 x 2 + 2 x + 1 + 2 d x = ∫ 1 x + 1 2 + 2 2 d x = log x + 1 + x 2 + 2 x + 3 + C 2 ∵ ∫ 1 x 2 + a 2 d x = log x + x 2 + a 2 + C Now, I = I 1 + I 2 ∵ ∫ x + 2 x 2 + 2 x + 3 d x = x 2 + 2 x + 3 + log x + 1 + x 2 + 2 x + 3 + C , where C = C 1 + C 2

Evaluate ∫ e x ( x - 1 ) ( x + 1 ) 3 d x .

Let I = ∫ e x x - 1 ( x + 1 ) 3 d x ⇒ I = ∫ e x ( x + 1 ) - 2 ( 1 + x ) 3 d x ⇒ I = ∫ e x 1 ( 1 + x ) 2 - 2 ( 1 + x ) 3 d x Here, d d x 1 1 + x 2 = - 2 1 + x 3 We know that ∫ e x f ( x ) + f ' ( x ) d x = e x f x + C Hence, I = e x ( x + 1 ) 2 + C

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Nagaland Board-2019

Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Nagaland Board-2019

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Nagaland Board-2019

Attempt the free practice questions on Chapter 7: Integrals, Exercise 1: Nagaland Board-2019 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.