Question

\[\int\limits_0^{\pi/2} 2 \sin x \cos x \tan^{- 1} \left( \sin x \right) dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{2} 2 \sin x \cos x \tan^{- 1} \left( \sin x \right) d x . Then, \]
\[Let \sin x = t . Then, \cos x dx = dt\]
\[When\ x = 0, t = 0\ and\ x = \frac{\pi}{2}, t = 1\]
\[ \therefore I = \int_0^1 2t \tan^{- 1} t dt\]
\[ \Rightarrow I = 2 \left[ \frac{t^2 \tan^{- 1} t}{2} \right]_0^1 - 2 \int_0^1 \frac{t^2}{1 + t^2} dt\]
\[ \Rightarrow I = 2 \left[ \frac{t^2 \tan^{- 1} t}{2} \right]_0^1 - 2 \int_0^1 \left( \frac{1 + t^2}{1 + t^2} - \frac{1}{1 + t^2} \right) dt\]
\[ \Rightarrow I = 2 \left[ \frac{t^2 \tan^{- 1} t}{2} \right]_0^1 - \left[ t - \tan^{- 1} t + \right]_0^1 \]
\[ \Rightarrow I = 1 \tan^{- 1} 1 - 0 - 1 + \tan^{- 1} 1 + 0\]
\[ \Rightarrow I = \frac{\pi}{4} - 1 + \frac{\pi}{4}\]
\[ \Rightarrow I = \frac{\pi}{2} - 1\]