Question

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

Answer 1

Solution : -

$I=\int ({\sin }^{-1}x{)}^{2}dx$

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

Put ${\sin }^{-1}x=t$ ___(1)

$\sin t=x$ ____(2)

$\cos tdt=dx$ ____(3)

$I=\int t^2\cos tdt$

$I=t^2\int \cos tdt-\int 2t\sin tdt$

$=t^2\sin t-2[\int t.\sin tdt]$

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

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

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