Question

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

Answer 1

Let $x=tan\theta$

$dx=se{c}^2\theta d\theta$

$\frac{2tan\theta }{1+ta{n}^2\theta }=sin\frac{\theta }{2}$

$\int si{n}^{-1}\left(sin\frac{\theta }{2}\right)se{c}^2\theta d\theta$

$\int \frac{\theta }{2}se{c}^2\theta d\theta$

Integrating by parts, we get

$u=\frac{\theta }{2},v=se{c}^2\theta$

$du=\frac{1}{2},\int vd\theta =tan\theta$

$\frac{\theta }{2}tan\theta -\frac{1}{2}\int tan\theta d\theta$

$\frac{xta{n}^{-1}x}{2}+\frac{log|cosx|}{2}+C$