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

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

Evaluate : ∫ 0 a sin - 1 x a + x d x

Let, ∫ 0 a sin - 1 x a + x d x Let x = a tan 2 t Differentiating both the sides, we get d x = 2 a tan t sec 2 t d t When x = 0 , then a tan 2 t = 0 ⇒ t = 0 When x = a , then a tan 2 t = a ⇒ tan t = 1 ⇒ t = π 4 ∴ I = ∫ 0 π 4 sin - 1 a tan 2 t a + a tan 2 t 2 a tan t sec 2 t d t = ∫ 0 π 4 sin - 1 a tan 2 t a sec 2 t 2 a tan t sec 2 t d t ∵ sec 2 x = 1 + tan 2 x = ∫ 0 π 4 sin - 1 sin 2 t 2 a tan t sec 2 t d t = ∫ 0 π 4 sin - 1 sin t 2 a tan t sec 2 t d t = ∫ 0 π 4 t 2 a tan t sec 2 t d t = 2 a ∫ 0 π 4 t tan t sec 2 t d t Consider, ∫ t tan t sec 2 t d t = t ∫ sec t sec t tan t d t - ∫ d d t t ∫ sec t sec t tan t d t d t [Integration by parts] = t sec 2 t 2 - ∫ 1 sec 2 t 2 d t ∵ ∫ sec t sec t tan t d t = ∫ z d z = z 2 2 , where z = sec t = t sec 2 t 2 - tan t 2 So, I = 2 a ∫ 0 π 4 t tan t sec 2 t d t = 2 a t sec 2 t 2 - tan t 2 0 π 4 = 2 a π 4 sec 2 π 4 2 - tan π 4 2 - 0 sec 2 0 2 - tan 0 2 = 2 a π 4 2 2 - 1 2 - 0 - 0 ∴ I = a π 2 - 1 .

Evaluate: ∫ 1 2 log x 2 d x

To evaluate ∫ 1 2 log x 2 d x . We have, ∫ 1 · log x 2 d x = log x 2 ∫ 1 d x - ∫ d d x log x 2 ∫ 1 d x d x [Using integration by parts] = log x 2 x - ∫ 2 log x · 1 x x d x = x log x 2 - 2 ∫ log x d x = x log x 2 - 2 log x ∫ 1 d x - ∫ d d x log x ∫ 1 d x d x [Using integration by parts] = x log x 2 - 2 log x x - ∫ 1 x x d x = x log x 2 - 2 x log x - x So, ∫ 1 2 log x 2 d x = x log x 2 - 2 x log x - x 1 2 = 2 log 2 2 - 2 2 log 2 - 2 - 1 log 1 2 - 2 1 log 1 - 1 = 2 log 2 2 - 2 log 2 + 2 - 0 - 2 0 - 1 ∴ ∫ 1 2 log x 2 d x = 2 log 2 2 - 2 log 2 + 1 .

Evaluate: ∫ 0 1 2 x sin - 1 x 1 - x 2 d x

Given, ∫ 0 1 2 x sin - 1 x 1 - x 2 d x Let t = sin - 1 x ⇒ x = sin t d t = 1 1 - x 2 d x When x = 0 , t = sin - 1 0 = 0 When x = 1 2 , t = sin - 1 1 2 = π 6 So, ∫ 0 1 2 x sin - 1 x 1 - x 2 d x = ∫ 0 π 6 t sin t d t We have, ∫ t sin t d t = t ∫ sin t d t - ∫ d ( t ) d t ∫ sin t d t d t [Using integration by parts] = - t cos t + ∫ cos t d t = - t cos t + sin t So, ∫ 0 π 6 t sin t d t = - t cos t + sin t 0 π 6 = - π 6 cos π 6 + sin π 6 - - 0 cos 0 + sin 0 = - π 6 3 2 + 1 2 - 0 ∴ ∫ 0 1 2 x sin - 1 x 1 - x 2 d x = 1 2 - 3 π 12 .