Question

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

Answer 1

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