Integrate the function: $(\sin^{-1} x)^2$

Let $I = \int (\sin^{-1} x)^2 \cdot 1 \, dx$.
Using integration by parts,where $u = (\sin^{-1} x)^2$ and $dv = dx$,we have $du = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}} \, dx$ and $v = x$.
$I = x(\sin^{-1} x)^2 - \int x \cdot \frac{2 \sin^{-1} x}{\sqrt{1-x^2}} \, dx$.
$= x(\sin^{-1} x)^2 + \int \sin^{-1} x \cdot \left( \frac{-2x}{\sqrt{1-x^2}} \right) \, dx$.
Now,apply integration by parts again for the second integral,where $u = \sin^{-1} x$ and $dv = \frac{-2x}{\sqrt{1-x^2}} \, dx$.
Then $du = \frac{1}{\sqrt{1-x^2}} \, dx$ and $v = 2\sqrt{1-x^2}$.
$I = x(\sin^{-1} x)^2 + \left[ \sin^{-1} x \cdot 2\sqrt{1-x^2} - \int \frac{1}{\sqrt{1-x^2}} \cdot 2\sqrt{1-x^2} \, dx \right]$.
$= x(\sin^{-1} x)^2 + 2\sqrt{1-x^2} \sin^{-1} x - \int 2 \, dx$.
$= x(\sin^{-1} x)^2 + 2\sqrt{1-x^2} \sin^{-1} x - 2x + C$,where $C$ is the constant of integration.