Evaluate the following integrals:

Let $I=\int x^{2} \sin ^{-1} x d x$

Using integration by parts,

$I=\sin ^{-1} x \int x^{2} d x-\int \frac{d}{d x} \sin ^{-1} x \int x^{2} d x$

$=\frac{x^{3}}{3} \sin ^{-1} x-\int \frac{x^{3}}{3 \sqrt{1-x^{2}}} d x$

$I=\frac{x^{3}}{3} \sin ^{-1} x-\int I_{1}+C$        .......(i)

$\mathrm{I}_{1}=-\int \frac{\mathrm{x}^{3}}{3 \sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}$

Let $1-x^{2}=t^{2}$

$-2 x d x=2 t d t$

$-x d x=t d t$

$\mathrm{I}_{1}=-\int \frac{\left(1-\mathrm{t}^{2}\right) \mathrm{tdt}}{\mathrm{t}}$

$\mathrm{I}_{1}=\int\left(\mathrm{t}^{2}-1\right) \mathrm{dt}$

$=\frac{\mathrm{t}^{3}}{3}-\mathrm{t}+\mathrm{c}_{2}$

$=\frac{\left(1-x^{2}\right)^{\frac{3}{2}}}{3}-\left(1-x^{2}\right)^{\frac{1}{2}}+c_{2}$

$=\frac{x^{3}}{3} \sin ^{-1} x-\frac{\left(1-x^{2}\right)^{\frac{3}{2}}}{9}+\frac{1}{3}\left(1-x^{2}\right)^{\frac{1}{2}}+c$