Question

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

Answer 1

$Puttingx=\sin tanddx=\cos tdt,$

$weget$

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

$=t^2.\left(\sin t\right)-\int 2t\left(\sin t\right)dt$

$=t^2\sin t-2\left[t\left(-\cos t\right)-\int 1.\left(-\cos t\right)dt\right]$

$=t^2\sin t+2t\cos t-2\sin t+C$

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