Put $\sin ^{-1} x=\theta \Rightarrow x=\sin \theta \Rightarrow d x=\cos \theta \mathrm{d} \theta$
$\begin{aligned}
&\therefore \int\left(\sin ^{-1} x\right)^2 d x=\int \theta^2 \cos \theta d \theta \\
&=\theta^2 \sin \theta+2 \theta \cos \theta-2 \int \cos \theta d \theta+C \\
&=\theta^2 \sin \theta+2 \theta \sqrt{1-\sin ^2} \theta-2 \sin \theta+C \\
&=x\left(\sin ^{-1} x\right)^2+2 \sin ^{-1} x \cdot \sqrt{1-x^2}-2 x+C
\end{aligned}$