Solution:

Let $\sin ^{-1} \mathrm{x}=\mathrm{t}$
$\begin{array}{l}
\Rightarrow \mathrm{d}\left(\sin ^{-1} \mathrm{x}\right)=\mathrm{d} \mathrm{t} \\
\Rightarrow \frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^{2}}}=\mathrm{dt}
\end{array}$
$\therefore$ By substituting $\mathrm{t}$ and $\mathrm{dt}$ in the given equation we obtain
$\begin{array}{l}
\Rightarrow \int \frac{1}{t^{2}} d t \\
\Rightarrow \int t^{-2} \cdot d t
\end{array}$
Now, on integrating the above equation we obtain
$\Rightarrow \frac{\mathrm{t}^{-1}}{-1}+\mathrm{C}$
But $t=\sin ^{-1} x$
$\Rightarrow \frac{-1}{\sin ^{-1} x}+c$