Find the value of ∫ d x 3 x 2 + 13 x - 10 .

Let the given integral is I = ∫ d x 3 x 2 + 13 x - 10 = ∫ d x 3 x 2 + 13 3 x - 10 3 = d x 3 x 2 + 13 3 x + 13 6 2 - 13 6 2 - 10 3 = ∫ d x 3 x + 13 6 2 - 169 36 + 10 3 = 1 - 3 ∫ d x x + 13 6 2 - 17 6 2 = 1 3 × 1 2 × 17 4 log x + 13 6 - 17 6 x + 13 6 + 17 6 + c = 1 17 log x - 2 3 x + 5 + c

Find the value of ∫ e tan - 1 x 1 + x 2 d x .

Let I = e tan - 1 x 1 + x 2 Put tan - 1 x = t 1 1 + x 2 = d t d x d x = 1 + x 2 d t Now, I = ∫ e t 1 + x 2 · 1 + x 2 d t = ∫ e t · d t = e t + C Thus, I = e tan - 1 x + C

Find the value of ∫ 1 - sin x cos 2 x d x .

Given, ∫ 1 - sin x cos 2 x d x = ∫ 1 cos 2 x - sin x cos 2 x d x ∫ 1 cos 2 x - sin x cos x × 1 cos x d x = ∫ sec 2 x - sec x tan x d x = tan x - sec x + C

Find the value of ∫ 0 π / 4 sin 2 x d x .

Given that, ∫ 0 π / 4 sin 2 x d x Let 2 x = t Differentiating on both sides we get 2 d x = d t ⇒ 1 2 ∫ 0 π / 4 sin t · d t = - cos ( t ) 4 0 π / 4 = - cos ( 2 x ) 4 0 π / 4 = 0 . 5

Find the value of ∫ sin x ( 1 + cos x ) 2 d x .

Given that, the integral is ∫ sin x ( 1 + cos x ) 2 d x Let 1 + cos x = t ⇒ - sin x d x = d t ⇒ ∫ sin x ( 1 + cos x ) 2 d x = ∫ - d t ( t ) 2 = - t - 2 + 1 - 2 + 1 + c = 1 t + c = 1 1 + cos x + c

Find the value of ∫ e 3 x · cos 2 x d x .

Let the integral be I = ∫ e 3 x · cos 2 x d x We know that, ∫ f x · g x d x = f x ∫ g x d x - ∫ f ' x ∫ g x d x ⇒ ∫ e 3 x · cos 2 x d x = cos 2 x ∫ e 3 x d x - ∫ d d x cos 2 x ∫ e 3 x d x ⇒ I = cos 2 x · e 3 x 3 - ∫ - 2 sin 2 x · e 3 x 3 d x ⇒ I = cos 2 x e 3 x 3 + 2 3 ∫ sin 2 x e 3 x d x ⇒ I = cos 2 x e 3 x 3 + 2 3 sin 2 x · e 3 x 3 - ∫ d d x ( sin 2 x ) ∫ e 3 x d x d x I = cos 2 x e 3 x 3 + 2 3 sin 2 x e 3 x 3 - ∫ 2 cos 2 x e 3 x 3 d x I = cos 2 x · e 3 x 3 + 2 3 sin 2 x · e 3 x 3 - 2 3 I I = cos 2 x · e 3 x 3 + 2 3 sin 2 x · e 3 x 3 - 4 9 I I + 4 9 I = cos 2 x e 3 x 3 + 2 9 sin 2 x · e 3 x 13 9 I = 3 cos 2 x e 3 x + 2 sin 2 x e 3 x 9 I = 1 13 3 cos 2 x · e 3 x + 2 sin 2 x · e 3 x + C

Find the value of ∫ 0 8 a x x + 8 a - x d x .

Let I = ∫ 0 8 a x d x x + 8 a - x ....(i) From the property of definite integral ∫ 0 a F ( x ) d x = ∫ 0 a F ( a - x ) d x Now, I = ∫ 0 8 a ( 8 a - x ) 8 a - x + 8 a - ( 8 a - x ) = ∫ 0 8 a ( 8 a - x ) d x 8 a - x + x ........(ii) Add equation(i), equation (ii), we get 2 I = ∫ 0 8 a x x + 8 a - x d x + ∫ 0 8 a ( 8 a - x ) 8 a - x + x d x = ∫ 0 8 a x + 8 a - x x + 8 a - x d x ⇒ 2 I = ∫ 0 a d x ⇒ 2 I = a - 0 ⇒ I = a 2 Thus, the value of ∫ 0 8 a x x + 8 a - x d x = a 2

Prove that: ∫ π / 6 π / 3 d x 1 + tan x = π 12

Let y = ∫ π 6 π 3 cos x sin x + cos x d x . . . . 1 Recall that, ∫ a b f ( x ) d x = ∫ a b f ( a + b - x ) d x Therefore, we get, y = ∫ π 6 π 3 cos π 3 + π 6 - x sin π 3 + π 6 - x + cos π 3 + π 6 - x d x ⇒ y = ∫ π 6 π 3 sin x ( cos x + sin x ) d x . . . ( 2 ) Adding eq. ( 1 ) and eq . ( 2 ) 2 y = ∫ π 6 π 3 1 d x 2 y = ( x ) π 6 π 3 y = π 12 Hence, proved.

E M B I B E

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

Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Chhattisgarh Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Integrals, Exercise 1: Chhattisgarh Board-2018

Attempt the free practice questions on Chapter 7: Integrals, Exercise 1: Chhattisgarh Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.