Question

$\underset{0}{\overset{1}{\int }}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$

Answer 1

$\begin{array}{cc}\text{Let}x=\tan \theta & ⟹\theta ={\tan }^{-1}x\\ & ⟹{\sin }^{-1}\frac{2x}{1+x^2}={\sin }^{-1}\frac{2\tan \theta }{1+{\tan }^2\theta }\\ & ={\sin }^{-1}\left(\sin 2\theta \right)\text{using}\left(1\right)\\ & =2\theta =2{\tan }^{-1}x\end{array}$

$\begin{array}{c}\text{Now the integral becomes fairly simple}\\ I=2{\int }_1^0{\tan }^{-1}x\cdot 1dx\\ \text{Integrating by parts}\\ \int u\cdot vdx=u\int vdx-\int (\frac{du}{dx}\cdot \int vdx)dx\end{array}$

$\begin{array}{c}\text{Taking}v=1\&u={\tan }^{-1}x\\ ⟹I=x\cdot {\tan }^{-1}{x|}_0^1-{\frac{\ln |1+x^2|}{2}|}_0^1\end{array}$

$⟹I=\frac{\pi }{4}-\frac{\ln 2}{2}$