Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\sin 2x{\tan }^{-1}\left(\sin x\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2x{\tan }^{-1}\left(\sin x\right)\mathit{}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}\mathit{,} \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}2\sin x\cos x{\tan }^{-1}\left(\sin x\right)\mathit{}\mathit{d}x \\ \mathrm{Let}\sin x=t.\mathrm{Then},\cos xdx=dt \\ \mathrm{When}x=0,t=0\mathrm{and}x=\frac{\mathrm{\pi }}{2},t=1 \\ \therefore I=2{\int }_0^{1}t{\tan }^{-1}tdt \\ \Rightarrow I=2{\left[\frac{t^2}{2}{\tan }^{-1}t\right]}_0^1-2{\int }_0^1\frac{t}{1+t^2}dt \\ \Rightarrow I=2{\left[\frac{t^2}{2}{\tan }^{-1}t\right]}_0^1-{\left[\log \left(1+t^2\right)\right]}_0^1 \\ \Rightarrow I=\frac{2\mathrm{\pi }}{4}-1 \\ \Rightarrow I=\frac{\mathrm{\pi }}{2}-1 \end{gathered}$$