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

: Let $I = \int\limits_0^{\pi /2} {\sin 2x{{\tan }^{ - 1}}\left( {\sin x} \right)dx}$

$\Rightarrow$ $I = \int\limits_0^{\pi /2} {2\sin x\cos x{{\tan }^{ - 1}}\left( {\sin x} \right)dx}$

Put $\sin x = t$ $\Rightarrow$ $\cos xdx = dt$

When $x = 0,t = 0$ and when $x = \cfrac{\pi }{2},t = 1$

$\therefore$ $I = 2\int\limits_0^1 {t{{\tan }^{ - 1}}t\,dt = 2\left[ {{{\tan }^{ - 1}}t\int t dt - \int {\left( {\cfrac{d}{{dt}}\left( {{{\tan }^{ - 1}}t} \right) \cdot \int t dt} \right)} dt} \right]} _0^1$

$= 2\left[ {{{\tan }^{ - 1}}\left( t \right)\cfrac{{{t^2}}}{2} - \int {\cfrac{1}{{1 + {t^2}}} \cdot \cfrac{{{t^2}}}{2}} dt} \right]_0^1$
$= 2\left[ {\cfrac{{{t^2}}}{2}{{\tan }^{ - 1}}\left( t \right) - \cfrac{1}{2}\int {\cfrac{{1 + {t^2} - 1}}{{1 + {t^2}}} \cdot } dt} \right]_0^1$]

$= \left[ {{t^2}{{\tan }^{ - 1}}\left( t \right) - \int {\left( {1 - \cfrac{1}{{1 + {t^2}}}} \right)dt} } \right]_0^1 = \left[ {{t^2}{{\tan }^{ - 1}}\left( t \right) - t + {{\tan }^{ - 1}}t} \right]_0^1$

$= {\tan ^{ - 1}}\left( 1 \right) - 1 + {\tan ^{ - 1}}1 = \cfrac{\pi }{4} - 1 + \cfrac{\pi }{4} = \cfrac{\pi }{2} - 1$