Find ∫ 1 5 x - 6 - x 2 d x

Given, ∫ 1 5 x - 6 - x 2 d x Consider, 5 x - 6 - x 2 = - x 2 - 5 x + 6 = - x 2 - 5 x + 6 + 25 4 - 25 4 = - x 2 - 5 x + 25 4 + 6 - 25 4 = - x 2 - 5 x + 5 2 2 + 24 - 25 4 = - x - 5 2 2 - 1 4 ∵ a 2 - 2 a b + b 2 = a - b 2 = 1 4 - x - 5 2 2 = 1 2 2 - x - 5 2 2 ∴ ∫ 1 5 x - 6 - x 2 d x = ∫ 1 1 2 2 - x - 5 2 2 d x = sin - 1 x - 5 2 1 2 + C [Using ∫ 1 a 2 - x 2 d x = sin - 1 x a + C ] = sin - 1 2 x - 5 2 1 2 + C = sin - 1 2 x - 5 + C Hence, ∫ 1 5 x - 6 - x 2 d x = sin - 1 2 x - 5 + C

Find ∫ d x x 6 log x 2 + 7 log x + 2

Given ∫ d x x 6 log x 2 + 7 log x + 2 Let u = log x Differentiate with respect to x , we get d u d x = 1 x ⇒ d u = d x x ∴ ∫ d x x 6 log x 2 + 7 log x + 2 = ∫ d u 6 u 2 + 7 u + 2 Consider, 6 u 2 + 7 u + 2 = 6 u 2 + 7 6 u + 2 6 = 6 u 2 + 7 6 u + 1 3 + 49 144 - 49 144 = 6 u 2 + 7 6 u + 49 144 + 1 3 - 49 144 = 6 u 2 + 7 6 u + 7 12 2 + 48 - 49 144 = 6 u + 7 12 2 - 1 144 = 6 u + 7 12 2 - 1 12 2 ∵ a + b 2 = a 2 + 2 a b + b 2 ∴ ∫ d u 6 u 2 + 7 u + 2 = 1 6 ∫ 1 u + 7 12 2 - 1 12 2 We know that, ∫ 1 x 2 - a 2 = 1 2 a log x - a x + a + C = 1 6 × 1 2 1 12 log u + 7 12 - 1 12 u + 7 12 + 1 12 + C = 1 6 × 1 1 6 log 12 u + 7 - 1 12 12 u + 7 + 1 12 + C = log 2 6 u + 3 2 6 u + 8 + C = log 6 u + 3 6 u + 8 + C = log 6 log x + 3 6 log x + 8 + C [Substitute u = log x ] ∴ ∫ d x x 6 log x 2 + 7 log x + 2 = log 6 log x + 3 6 log x + 8 + C

Prove that ∫ 0 π log e 1 + cos x d x = πlog e 1 2 .

Let I = ∫ 0 π log ( 1 + cos x ) d x = ∫ 0 π log 2 cos 2 x 2 d x = ∫ 0 π log 2 + logcos 2 x 2 d x = ∫ 0 π ( log 2 ) d x + ∫ 0 π 2 logcos x 2 d x = π log 2 + 2 ∫ 0 π logcos x 2 d x = π log 2 + 2 l 1 . . . . . . ( 1 ) In I 1 , Put x 2 = t so that d x = 2 d t I 1 = ∫ 0 π 2 logcos t · 2 d t = 2 ∫ 0 π 2 logcos x d x . . . . i We know that ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x Therefore, I 1 = 2 ∫ 0 π 2 logcos π 2 - x d x I 1 = 2 ∫ 0 π 2 log sin x d x . . . . i i Adding i and i i 2 I 1 = 2 ∫ 0 π 2 log ( sin x ) d x + 2 ∫ 0 π 2 log ( cos x ) d x I 1 = ∫ 0 π 2 log ( sin x ) + log ( cos x ) d x I 1 = ∫ 0 π 2 log ( sin x cos x ) d x I 1 = ∫ 0 π 2 log ( sin 2 x 2 ) d x I 1 = ∫ 0 π 2 logsin 2 x d x - ∫ 0 π 2 log 2 d x Put 2 x = t , d x = d t 2 I 1 = 1 2 ∫ 0 π logsin t d t - ∫ 0 π 2 log 2 d x I 1 = 1 2 ∫ 0 π logsin x d x - ∫ 0 π 2 log 2 d x We know that, ∫ 0 2 a f ( x ) d x = 2 ∫ 0 a f ( x ) d x , if f ( 2 a - x ) = f ( x ) I 1 = ∫ 0 π 2 logsin x d x - ∫ 0 π 2 log 2 d x I 1 = 1 2 I 1 - ∫ 0 π 2 log 2 d x 1 2 I 1 = - log ( 2 ) [ x ] 0 π 2 I 1 = - π 4 log ( 2 ) Hence, from equation 1 I = π log 2 + 4 - π 4 log 2 = π log 2 - 2 π log 2 = - π log 2 = π log e ( 1 2 )

Find the value of ∫ 0 π x sin x 1 + cos 2 x . d x

By the given equation I = ∫ 0 π x sin x d x 1 + cos 2 x = ∫ 0 π π - x sin π - x 1 + cos 2 π - x d x = ∫ 0 π π - x sin x 1 + cos 2 x d x = ∫ 0 π πsin x 1 + cos 2 x d x - ∫ 0 π x sin x 1 + cos 2 x d x 2 I = π ∫ 0 π sin x dx 1 + cos 2 x Put cos x = v - sin x d x = d v When x = 0 , v = 1 and x = π , v = - 1 2 I = π ∫ 1 - 1 - dv 1 + v 2 = - π tan - 1 v 1 - 1 2 I = - π tan - 1 [ - 1 ] + tan - 1 [ 1 ] I = - π 2 - π 4 - π 4 = π 2 4

Integrate 2 x + 5 x 2 + 3 x + 1 with respect to x .

Let I = ∫ 2 x + 5 x 2 + 3 x + 1 d x I = ∫ 2 x + 3 + 2 x 2 + 3 x + 1 d x I = ∫ 2 x + 3 x 2 + 3 x + 1 d x + ∫ 2 x 2 + 3 x + 1 d x . . . . 1 Let I = I 1 + I 2 Consider I 1 = ∫ 2 x + 3 x 2 + 3 x + 1 d x We know that ∫ f ' x f x d x = log f x + C Therefore, I 1 = log x 2 + 3 x + 1 + C 1 Consider I 2 = ∫ 2 x 2 + 3 x + 1 d x I 2 = ∫ 1 x + 3 2 2 - 5 4 d x I 2 = ∫ 1 x + 3 2 2 - 5 2 2 d x Put x + 3 2 = t then d x = d t I 2 = ∫ 1 t 2 - 5 2 2 dt We know that ∫ d x x 2 - a 2 = log x + x 2 - a 2 + C I 2 = log t + t 2 - 5 2 2 + C 2 Substitute I 1 and I 2 in equation 1 I = log x 2 + 3 x + 1 + C 1 + log t + t 2 - 5 2 2 + C 2 I = log x 2 + 3 x + 1 + log x + 3 2 + x + 3 2 2 - 5 2 2 + C

Integrate cos - 1 x a + x with respect to x .

I = ∫ cos − 1 ⁡ x a + x d x x = a tan 2 θ d x = 2 a tan ⁡ θ sec 2 ⁡ θ d θ I = ∫ cos − 1 ⁡ a tan 2 ⁡ θ a + a tan 2 ⁡ θ ⋅ 2 a tan θ sec 2 θ d θ I = ∫ cos − 1 ⁡ tan 2 ⁡ θ sec 2 ⁡ θ ⋅ 2 a tan θ sec 2 θ d θ I = ∫ cos − 1 sin θ ⋅ 2 a tan θ sec 2 θ d θ I = 2 a ∫ cos - 1 [ cos ( π 2 - θ ) ] tan θ sec 2 θ d θ I = 2 a ∫ ( π 2 - θ ) tan θ sec 2 θ d θ I = a π ∫ tan θ sec 2 θ d θ - 2 a ∫ θ tan θ sec 2 θ d θ . . . 1 Let I = I 1 + I 2 Consider I 1 = a π ∫ tan θ sec 2 θ d θ Put tan θ = t then sec 2 θ d θ = d t Therefore, I 1 = a π ∫ t d t I 1 = a π t 2 2 + C I 1 = a π tan 2 θ 2 + C 1 Consider I 2 = - 2 a ∫ θ tan θ sec 2 θ d θ We know that ∫ f g ′ = f g − ∫ f ' g f = θ , g ' = sec 2 ( θ ) tan ( θ ) and f ' = 1 , g = sec 2 ( θ ) 2 I 2 = - 2 a θ sec 2 ( θ ) 2 - ∫ sec 2 ( θ ) 2 d θ I 2 = - a θ sec 2 ( θ ) - tan ( θ ) + C 2 Substitute I 1 and I 2 in equation 1 , we get I = a π tan 2 θ 2 + C 1 - a θ sec 2 ( θ ) - tan ( θ ) + C 2 I = a π tan 2 θ 2 - a θ sec 2 ( θ ) - tan ( θ ) + C , where θ = tan - 1 x a

Prove that: I = ∫ log ( 1 + cos x ) d x = πlog e 1 2

Let I = ∫ 0 π log ( 1 + cos x ) d x = ∫ 0 π log 2 cos 2 x 2 d x = ∫ 0 π log 2 + logcos 2 x 2 d x = ∫ 0 π ( log 2 ) d x + ∫ 0 π 2 logcos x 2 d x = π log 2 + 2 ∫ 0 π logcos x 2 d x = π log 2 + 2 l 1 . . . . . . ( 1 ) In I 1 , Put x 2 = t so that d x = 2 d t I 1 = ∫ 0 π 2 logcos t · 2 d t = 2 ∫ 0 π 2 logcos x d x . . . . i We know that ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x Therefore, I 1 = 2 ∫ 0 π 2 logcos π 2 - x d x I 1 = 2 ∫ 0 π 2 log sin x d x . . . . i i Adding i and i i 2 I 1 = 2 ∫ 0 π 2 log ( sin x ) d x + 2 ∫ 0 π 2 log ( cos x ) d x I 1 = ∫ 0 π 2 log ( sin x ) + log ( cos x ) d x I 1 = ∫ 0 π 2 log ( sin x cos x ) d x I 1 = ∫ 0 π 2 log ( sin 2 x 2 ) d x I 1 = ∫ 0 π 2 logsin 2 x d x - ∫ 0 π 2 log 2 d x Put 2 x = t , d x = d t 2 I 1 = 1 2 ∫ 0 π logsin t d t - ∫ 0 π 2 log 2 d x I 1 = 1 2 ∫ 0 π logsin x d x - ∫ 0 π 2 log 2 d x We know that, ∫ 0 2 a f ( x ) d x = 2 ∫ 0 a f ( x ) d x , if f ( 2 a - x ) = f ( x ) I 1 = ∫ 0 π 2 logsin x d x - ∫ 0 π 2 log 2 d x I 1 = 1 2 I 1 - ∫ 0 π 2 log 2 d x 1 2 I 1 = - log ( 2 ) [ x ] 0 π 2 I 1 = - π 4 log ( 2 ) Hence, from equation 1 I = π log 2 + 4 - π 4 log 2 = π log 2 - 2 π log 2 = - π log 2 = π log e ( 1 2 )

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 7 - Integrals>Rajasthan Board-2018

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

Author:Embibe Experts

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

Attempt the free practice questions on Chapter 7: Integrals, Exercise 1: Rajasthan 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.