Answer
$$x(\sin^{-1}x)^2+2(\sin^{-1}x)\sqrt{1-x^2}-2x+c$$
Work Step by Step
Given $$ \int(\sin^{-1}x)^2dx$$
Let
\begin{align*}
u&=(\sin^{-1}x)^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=dx\\
du&=2(\sin^{-1}x) \frac{dx}{\sqrt{1-x^2}}\ \ \ \ \ \ \ \ \ \ v=x
\end{align*}
Then using integration by parts
\begin{align*}
\int(\sin^{-1}x)^2dx&=uv-\int vdu\\
&=x(\sin^{-1}x)^2- \int (\sin^{-1}x) \frac{2xdx}{\sqrt{1-x^2}}
\end{align*}
Let
\begin{align*}
u&=(\sin^{-1}x) \ \ \ \ \ \ \ \ \ \ \ \ dv=\frac{2xdx}{\sqrt{1-x^2}}\\
du&= \frac{dx}{\sqrt{1-x^2}}\ \ \ \ \ \ \ \ \ \ \ \ \ v=-2\sqrt{1-x^2}
\end{align*}
\begin{align*}
\int (\sin^{-1}x) \frac{2xdx}{\sqrt{1-x^2}}&=-2(\sin^{-1}x)\sqrt{1-x^2}+2\int dx\\
&= -2(\sin^{-1}x)\sqrt{1-x^2}+2x
\end{align*}
Hence
\begin{align*}
\int(\sin^{-1}x)^2dx &=x(\sin^{-1}x)^2- \int (\sin^{-1}x) \frac{2xdx}{\sqrt{1-x^2}}\\
&=x(\sin^{-1}x)^2+2(\sin^{-1}x)\sqrt{1-x^2}-2x+c
\end{align*}
