Evaluate the following integrals:

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

we know that, $\cos ^{3} t d t=\frac{3 \cos t+\cos 3 t}{4}$

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

$=\frac{1}{4} \int x(3 \cos x+\cos 3 x) d x$

Using integration by parts,

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

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

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

$I=\frac{3 x \sin x}{4}+\frac{x \sin x}{12}+\frac{3 \cos x}{4}+\frac{\cos 3 x}{36}+c$