Evaluate: $\int_{0}^{\pi / 2} 2 \sin x \cos x \tan ^{-1}(\sin x) d x$

CBSE Class 12 · 2011 · Definite IntegralsLong Answer

Evaluate:

\int_{0}^{π /2} 2 sin x cos x tan^{- 1} (sin x) d x

Answer: $\frac{π}{2} - 1$

Put

z = sin x \Rightarrow d z = cos x d x

;

x = 0 \to z = 0

x = \frac{π}{2} \to z = 1

. Then

2 sin x cos x d x = 2 z d z

I = 2 \int_{0}^{1} z tan^{- 1} z d z

Integrate by parts with

u = tan^{- 1} z, d v = z d z

I = 2 [tan^{- 1} z \cdot \frac{z ^{2}}{2}]_{0}^{1} - \int_{0}^{1} \frac{z ^{2}}{1 + z ^{2}} d z = \frac{π}{4} - \int_{0}^{1} (1 - \frac{1}{1 + z ^{2}}) d z

= \frac{π}{4} - [z - tan^{- 1} z]_{0}^{1} = \frac{π}{4} - (1 - \frac{π}{4}) = \frac{π}{2} - 1

CBSE Class 12 Board · 2011 · Subjective