Question
Evaluate: $\int\limits_0^{\pi/2}$ 2 sin x cos x tan-1 (sin x) dx.
$ =\Bigg[\tan^{-1} t\times\text{t}^{2}\Bigg]_0^1-\int\limits_0^{1}\frac{{t}^{2}}{\text{t}^{2}+1}\text{dt}$
$=\frac{\pi}{4}-\int_0^1\Bigg(1-\frac{1}{\text{t}^{1}+1}\Bigg)\text{dt}$
$=\frac{\pi}{4}-[\text{t - tan}^{-1}\text{t}]^1_0$
$=\frac{\pi}{4}-\Bigg[1-\frac{\pi}{4}\Bigg]=\Bigg(\frac{\pi}{2}-1\Bigg)$.
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