Prove that $\int_{0}^{1} \sin^{-1} x \, dx = \frac{\pi}{2} - 1$.

(N/A) Let $I = \int_{0}^{1} \sin^{-1} x \cdot 1 \, dx$.
Using integration by parts,where $u = \sin^{-1} x$ and $dv = dx$:
$I = [x \sin^{-1} x]_{0}^{1} - \int_{0}^{1} \frac{x}{\sqrt{1-x^{2}}} \, dx$.
For the second integral,let $1 - x^{2} = t$,then $-2x \, dx = dt$,or $x \, dx = -\frac{1}{2} dt$.
When $x = 0, t = 1$ and when $x = 1, t = 0$.
$I = [x \sin^{-1} x]_{0}^{1} - \int_{1}^{0} \frac{1}{\sqrt{t}} \left(-\frac{1}{2}\right) \, dt$.
$I = [x \sin^{-1} x]_{0}^{1} - \frac{1}{2} \int_{0}^{1} t^{-1/2} \, dt$.
$I = [x \sin^{-1} x]_{0}^{1} - \frac{1}{2} [2\sqrt{t}]_{0}^{1}$.
$I = (1 \cdot \sin^{-1}(1) - 0 \cdot \sin^{-1}(0)) - [\sqrt{t}]_{0}^{1}$.
$I = \frac{\pi}{2} - (\sqrt{1} - \sqrt{0}) = \frac{\pi}{2} - 1$.
Hence,the result is proved.