Evaluate $\int\limits_{0}^{\pi/2} \sin 2x \tan^{-1}(\sin x) dx.$

$\frac{\pi}{2} - 1$

The correct answer is Option (1) → $\frac{\pi}{2} - 1$

$I = \int\limits_{0}^{\pi/2} \sin 2x \tan^{-1}(\sin x) dx$

$I = \int\limits_{0}^{\pi/2} 2 \sin x \cos x \tan^{-1}(\sin x) dx$

Let $\sin x = t$, $\cos x dx = dt$

When $x = 0, t = 0$; when $x = \frac{\pi}{2}, t = 1$

$∴I = \int\limits_{0}^{1} 2t \tan^{-1} t dt$

$= 2 \left[ \tan^{-1} t \int t dt - \int \left( \frac{d}{dx} \tan^{-1} t \int t dt \right) dt \right]$

$= 2 \left[ \frac{t^2}{2} \tan^{-1} t - \int \frac{t^2}{2(1+t^2)} dt \right]_{0}^{1}$

$= \left[ t^2 \tan^{-1} t \right]_{0}^{1} - \int\limits_{0}^{1} \frac{t^2}{(1+t^2)} dt$

$= \tan^{-1} 1 - \left[ \int\limits_{0}^{1} 1 dt - \int\limits_{0}^{1} \frac{1}{1+t^2} dt \right]$

$= \frac{\pi}{4} - \left[ t - \tan^{-1} t \right]_{0}^{1}$

$= \frac{\pi}{4} - 1 + \frac{\pi}{4} = \frac{\pi}{2} - 1$

$∴\int\limits_{0}^{\pi/2} \sin 2x \tan^{-1}(\sin x) dx = \frac{\pi}{2} - 1$