Evaluate the following integrals:

Assume $\sin ^{-1} x^{2}=t$

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

$\Rightarrow \frac{2 \mathrm{xdx}}{\sqrt{1-\mathrm{x}^{4}}}=\mathrm{dt}$

$\Rightarrow \frac{\mathrm{xdx}}{\sqrt{1-\mathrm{x}^{4}}}=\frac{\mathrm{dt}}{2}$

$\therefore$ Substituting $t$ and dt in given equation we get

$\Rightarrow \int \frac{t}{2} d t$

$\Rightarrow \frac{1}{2} \int t . d t$

$\Rightarrow \frac{t^{2}}{4}+C$

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

$\Rightarrow \frac{\left(\sin ^{-1} x^{2}\right)^{2}}{4}+c$