Answer
$x(\sin^{-1} x)^2 +2 \sin^{-1} x(\sqrt {1-x^2}) -2x +C$
Work Step by Step
We will use integration by parts to obtain:
$I=\int (\sin^{-1} x)^2 \ dx \\= x (\sin^{-1} x)^2-2 \int \dfrac{x \sin^{-1} x}{\sqrt {1-x^2}} \ dx $
Plug in
$a=\sin^{-1} x \implies da=\dfrac{dx}{\sqrt {1-x^2}}$ and $x =\sin a$
Now,
$ x (\sin^{-1} x)^2-2 \int \dfrac{x \sin^{-1} x}{\sqrt {1-x^2}} \ dx =x(\sin^{-1} x)^2 -2 \int a \sin a \ da \\=x(\sin^{-1} x)^2 -2[a \cos a +\sin a +C] \\=x(\sin^{-1} x)^2 -\sin^{-1} x (\sqrt {1-x^2}) +x+C \\=x(\sin^{-1} x)^2 +2 \sin^{-1} x(\sqrt {1-x^2}) -2x +C$
