Evaluate the following integrals:

Let $\mathrm{I}=\int \mathrm{x} \sin ^{3} \mathrm{x} \mathrm{dx}$

We know that, $\sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}$

$=\int x\left(\frac{3 \sin x-\sin 3 x}{4}\right) d x$

$=\frac{1}{4} \int x(3 \sin x-\sin 3 x) d x$

Using integration by parts,

$I=\frac{1}{4}\left[x \int(3 \sin x-\sin 3 x) d x-\int 1 \int(3 \sin x-\sin 3 x) d x\right]$

$=\frac{1}{4}\left[x\left(-3 \cos x+\frac{\cos 3 x}{3}\right)-\int\left(-3 \cos x+\frac{\cos 3 x}{3}\right) d x\right]$

$=\frac{1}{4}\left[-3 x \cos x+\frac{x \cos 3 x}{3}+3 \sin x-\frac{\sin 3 x}{9}\right]+c$

$I=\frac{1}{36}[3 x \cos 3 x-27 x \cos x+27 \sin x-\sin 3 x]+c$