Question:
Evaluate the following integrals:
$\int \frac{x^{3} \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} d x$
Solution:
Let $I=\int \frac{x^{3} \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} d x$
$\sin ^{-1} x^{2}=t$
$\frac{1}{\sqrt{1-x^{4}}} 2 x d x=d t$
$I=\int \frac{x^{2} \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} x d x$
$=\int(\sin t) t \frac{d t}{2}$
Using integration by parts,
$=\frac{1}{2}\left[\mathrm{t} \int \sin t \mathrm{~d} t-\int \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t} \int \sin t \mathrm{dt}\right]$
$=\frac{1}{2}\left[-\mathrm{t} \cos t-\int-\mathrm{costdt}\right]$
$=\frac{1}{2}[-t \cos t+\sin t]+c$
$=\frac{1}{2}\left[x^{2}-\sqrt{1-x^{4}} \sin ^{-1} x^{2}\right]+c$