Evaluate ∫ 3 5 1 2 x + 3 - 2 x d x .

1 9 ( 13 ) 3 2 + ( 10 ) 3 2 - ( 9 ) 3 2 - ( 6 ) 3 2

Evaluate: ∫ 0 π 4 1 + sin 2 x . d x

Let I = ∫ 0 π 4 1 + sin 2 x . d x = ∫ 0 π 4 sin 2 x + cos 2 x + 2 sin x cos x . d x = ∫ 0 π 4 sin x + cos x 2 . d x ∴ a + b 2 = a 2 + b 2 + 2 a b = ∫ 0 π 4 sin x + cos x . d x = ∫ 0 π 4 sin x d x + ∫ 0 π 4 cos x . d x = - cos x 0 π 4 + sin x 0 π 4 = - cos π 4 - - cos 0 + sin π 4 - sin 0 = - 1 2 - - 1 + 1 2 - 0 = - 1 2 + 1 + 1 2 I = 1 Hence, the value of ∫ 0 π 4 1 + sin 2 x . d x = 1 .

Evaluate: ∫ 0 π 4 sin 4 x d x

Let I = ∫ 0 π 4 sin 4 x d x ⇒ I = ∫ 0 π 4 sin 2 x 2 d x ⇒ I = ∫ 0 π 4 1 - cos 2 x 2 2 d x ⇒ I = ∫ 0 π 4 1 - 2 cos 2 x + cos 2 2 x 4 d x ⇒ I = 1 4 ∫ 0 π 4 1 - 2 cos 2 x + cos 2 2 x d x ⇒ I = 1 4 ∫ 0 π 4 1 · d x - 1 2 ∫ 0 π 4 cos 2 x d x + 1 4 ∫ 0 π 4 cos 2 2 x d x ⇒ I = 1 4 [ x ] 0 π 4 - 1 2 sin 2 x 2 0 π 4 + 1 4 x 2 + sin 4 x 8 0 π 4 ⇒ I = 1 4 π 4 - 0 - 1 4 sin π 2 - sin 0 + 1 32 π + sinπ - 0 + sin 0 ⇒ I = π 16 - 1 4 1 - 0 + 1 32 π - 0 ⇒ I = π 16 - 1 4 + π 32 = 2 π - 8 + π 32 = 3 π - 8 32 Hence, ∫ 0 π 4 sin 4 x d x = 3 π - 8 32 .

Evaluate : ∫ 0 1 t 2 1 - t · d t

Let I = ∫ 0 1 t 2 1 - t · d t We use the property, ∫ 0 a f t . d t = ∫ 0 a f a - t . d t Let f t = t 2 1 - t Put t = 1 - t ∴ ∫ 0 1 t 2 1 - t . d t = ∫ 0 1 1 - t 2 1 - 1 + t . d t = ∫ 0 1 1 - 2 t + t 2 t . d t = ∫ 0 1 t 1 2 - 2 t . t 1 2 + t 2 . t 1 2 d t = ∫ 0 1 t 1 2 - 2 t 3 2 + t 5 2 d t = t 3 2 3 2 - 2 . t 5 2 5 2 + t 7 2 7 2 0 1 ∴ ∫ x n d x = x n + 1 n + 1 + c = 2 t 3 2 3 - 4 t 5 2 5 + 2 t 7 2 7 0 1 = 2 1 3 2 3 - 4 1 5 2 5 + 2 1 7 2 7 - 0 = 2 3 - 4 5 + 2 7 = 70 - 84 + 30 105 = 16 105 Hence, the value of ∫ 0 1 t 2 1 - t · d t = 16 105

Evaluate : ∫ 0 π x · sin x · cos 2 x · d x

Let I = ∫ 0 π x · sin x · cos 2 x · d x . . . . . . . . . ( 1 ) Using property, ∫ a b f x d x = ∫ a b f a + b - x d x ⇒ ∫ 0 π x · sin x · cos 2 x · d x = ∫ 0 π π - x sin π - x cos 2 π - x · d x As, sin π - x = sin x cos π - x = - cos x = ∫ 0 π π - x sin x cos 2 x · d x = ∫ 0 π πsin x cos 2 x · d x - ∫ 0 π x sin x cos 2 x · d x ⇒ I = ∫ 0 π πsin x cos 2 x · d x - I . . . by 1 ⇒ 2 I = ∫ 0 π πsin x cos 2 x · d x ⇒ 2 I = π ∫ 0 π sin x cos 2 x · d x Now, let cos x = t ∴ - sin x d x = d t When x = 0 , t = cos 0 = 1 When x = π , t = cos π = - 1 ⇒ 2 I = - π ∫ 1 - 1 t 2 · d t ⇒ 2 I = - π t 3 3 1 - 1 ⇒ 2 I = - π - 1 3 3 - 1 3 3 ⇒ 2 I = - π - 1 3 - 1 3 ⇒ 2 I = 2 π 3 ⇒ I = π 3 Hence, the value of ∫ 0 π x · sin x · cos 2 x · d x = π 3 .

Evaluate : ∫ 0 1 log x 1 - x 2 · d x

Let I = ∫ 0 1 log x 1 - x 2 · d x Let x = sin θ ∴ d x = cos θ d θ When x = 0 , x = sin θ = 0 ∴ θ = 0 x = 1 , x = sin θ = 1 ∴ θ = π 2 I = ∫ 0 π 2 log sin θ 1 - sin 2 θ · cos θ d θ I = ∫ 0 π 2 log sin θ cos θ · cos θ d θ I = ∫ 0 π 2 log sin θ d θ . . . . . . . . ( 1 ) Using property, ∫ 0 2 a f x d x = ∫ 0 a f x + f 2 a - x . d x Here, 2 a = π 2 ⇒ a = π 4 I = ∫ 0 π 4 log sin θ + log sin π 2 - θ . d θ I = ∫ 0 π 4 log sin θ + log cos θ . d θ I = ∫ 0 π 4 log sin θ cos θ . d θ ∴ log a b = log a + log b I = ∫ 0 π 4 log 2 sin θ cos θ 2 . d θ I = ∫ 0 π 4 log sin 2 θ 2 . d θ I = ∫ 0 π 4 log sin 2 θ - log 2 . d θ ∴ log a b = log a - log b I = ∫ 0 π 4 log sin 2 θ d θ - ∫ 0 π 4 log 2 . d θ I = I 1 - I 2 . . ( s a y ) I 2 = ∫ 0 π 4 log 2 . d θ I 2 = log 2 ∫ 0 π 4 . d θ I 2 = log 2 θ 0 π 4 I 2 = log 2 π 4 - 0 = π 4 log 2 Then, I 1 = ∫ 0 π 4 log sin 2 θ d θ Put 2 θ = t Then d θ = d t 2 When θ = 0 , t = 2 0 = 0 When θ = π 4 , t = 2 π 4 = π 2 ∴ I 1 = ∫ 0 π 2 log sin t × d t 2 ∴ I 1 = 1 2 ∫ 0 π 2 log sin t d t From 1 , ∴ I 1 = 1 2 I ∴ ∫ a b f x d x = ∫ a b f t d t Now, I = I 1 - I 2 I = 1 2 I - π 4 log 2 I - 1 2 I = - π 4 log 2 2 I - I 2 = - π 4 log 2 I 2 = - π 4 log 2 ∴ I = - π 2 log 2 = π 2 log 1 2 Hence, the value of I = ∫ 0 1 log x 1 - x 2 · d x = π 2 log 1 2 .

E M B I B E

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 4 - Definite Integration>EXERCISE 4.2

Maharashtra Board Solutions for Chapter: Definite Integration, Exercise 2: EXERCISE 4.2

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Definite Integration, Exercise 2: EXERCISE 4.2

Attempt the practice questions on Chapter 4: Definite Integration, Exercise 2: EXERCISE 4.2 with hints and solutions to strengthen your understanding. Mathematics & Statistics (Arts & Science) Part 2 Standard 12 solutions are prepared by Experienced Embibe Experts.