Question

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

Answer 1

$\begin{array}{cc}I=\int {\left({\sin }^{-1}x\right)}^{2}dx& \\ I=\int \left({\sin }^{-1}x\right).\left({\sin }^{-1}x\right)dx& \left[\begin{array}{c}Put{\sin }^{-1}=t\\ x=\sin t\\ dx=\cos tdt\end{array}\right]\\ =\int t^2.\cos tdt& \end{array}$

Using $ILATE$ formulae

$\begin{array}{c}=t^2\sin t-\int 2t\sin tdt\\ =t^2\sin t-2\int t\sin tdt\to ApplyILATEformulae\\ =t^2\sin t-2\left[-t\cos t+\int \cos tdt\right]\\ =t^2\sin t+2t\cos t-2\sin t+c\\ ={\left({\sin }^{-1}x\right)}^2\sin .{\sin }^{-1}x+2{\sin }^{-1}x\sqrt{1-x^2}-2\sin .{\sin }^{-1}x+c\\ ={\left({\sin }^{-1}x\right)}^2.x+2{\sin }^{-1}x\sqrt{1-x^2}-2x+c\end{array}$