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Let $I=\int_{0}^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x=\int_{0}^{\frac{\pi}{2}} 2 \sin x \cos x \tan ^{-1}(\sin x) d x$ Also, let $\sin x=t \Rightarrow \cos x d x=d t$

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

$\Rightarrow I=2 \int_{0}^{1} t \tan ^{-1}(t) d t$         ...(1)

Consider $\int t \cdot \tan ^{-1} t d t=\tan ^{-1} t \cdot \int t d t-\int\left\{\frac{d}{d t}\left(\tan ^{-1} t\right) \int t d t\right\} d t$

$=\tan ^{-1} t \cdot \frac{t^{2}}{2}-\int \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t$

$=\frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{2} \int \frac{t^{2}+1-1}{1+t^{2}} d t$

$=\frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{2} \int 1 d t+\frac{1}{2} \int \frac{1}{1+t^{2}} d t$

$=\frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{2} \cdot t+\frac{1}{2} \tan ^{-1} t$

$\Rightarrow \int_{0}^{1} t \cdot \tan ^{-1} t d t=\left[\frac{t^{2} \cdot \tan ^{-1} t}{2}-\frac{t}{2}+\frac{1}{2} \tan ^{-1} t\right]_{0}^{1}$

$=\frac{1}{2}\left[\frac{\pi}{4}-1+\frac{\pi}{4}\right]$

$=\frac{1}{2}\left[\frac{\pi}{2}-1\right]=\frac{\pi}{4}-\frac{1}{2}$

From equation (1), we obtain

$I=2\left[\frac{\pi}{4}-\frac{1}{2}\right]=\frac{\pi}{2}-1$