Question

Evaluate:
$\int ({\sin }^{-1}x{)}^2.dx$

Answer 1

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

$I=\int {\sin }^{-1}x.{\sin }^{-1}xdx$

Put ${\sin }^{-1}x=t;\sin t=x$

$\cos tdt=dx$

$I=\int t^2.\cos tdt=t^2.\sin t-2\int t\sin t\int dt=t^2\sin t-2[t\cos t+\int \cos tdt]$

$=t^2\sin t+2t\cos t-2\int \cos tdt=t^2\sin t+2\cos t-2\sin t+c$

$={\left({\sin }^{-1}x\right)}^2\sin \left({\sin }^{-1}x\right)+2{\sin }^{-1}x\sqrt{1-x^2}-2x+c={\left({\sin }^{-1}x\right)}^2.x+2{\sin }^{-1}x\sqrt{1-x^2}-2x+c$