Evaluate the following integrals:

Let

$\sqrt{x}=t$

$x=t^{2}$

$\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$

let $\mathrm{I}=2 \int \mathrm{t} \cos ^{3} \mathrm{tdt}$

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

$=2 \int \mathrm{t}\left(\frac{3 \cos t+\cos 3 t}{4}\right) \mathrm{dt}$

$=\frac{1}{2} \int \mathrm{t}(3 \cos t-\cos 3 \mathrm{t}) \mathrm{dt}$

Using integration by parts,

$=\frac{1}{2}\left[t\left(3 \sin t+\frac{1}{3} \sin 3 t\right)+\int\left(3 \sin t+\frac{\sin 3 t}{3}\right) d t\right]$

$=\frac{1}{2}\left[\frac{9 t \sin t+t \sin 3 t}{3}+\left\{3 \cos t+\frac{\cos 3 t}{9}\right\}\right]+c$

$=\frac{1}{18}[27 \mathrm{t} \sin \mathrm{t}+3 \mathrm{t} \sin 3 \mathrm{t}+9 \cos \mathrm{t}+\cos 3 \mathrm{t}]+\mathrm{c}$

$I=\frac{1}{18}[27 \sqrt{x} \sin \sqrt{x}+3 \sqrt{x} \sin 3 \sqrt{x}+9 \cos \sqrt{x}+\cos 3 \sqrt{x}]+c$