Evaluate the following integrals:

Assume $\sin ^{-1} x=t$

$\Rightarrow \mathrm{d}\left(\sin ^{-1} x\right)=\mathrm{dt}$

$\Rightarrow \frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^{2}}}=\mathrm{dt}$

$\therefore$ Substituting $t$ and $d t$ in the given equation we get

$\Rightarrow \int \frac{1}{\mathrm{t}^{2}} \mathrm{dt}$

$\Rightarrow \int \mathrm{t}^{-2} \cdot \mathrm{dt}$

$\Rightarrow \frac{\mathrm{t}^{-1}}{-1}+\mathrm{C}$

But $t=\sin ^{-1} x$

$\Rightarrow \frac{-1}{\sin ^{-1} x}+c$