Evaluate the following integrals:

Assume $\cos x=t$

$\Rightarrow \mathrm{d}(\cos x)=\mathrm{dt}$

$\Rightarrow-\sin x \mathrm{~d} x=\mathrm{dt}$

$\Rightarrow \mathrm{dx}=\frac{-\mathrm{dt}}{\sin \mathrm{x}}$

$\therefore$ Substituting $t$ and $d t$ in the given equation we get

$\Rightarrow \int \sqrt[3]{t^{2}} \sin x \cdot \frac{d t}{\sin x}$

$\Rightarrow \int t^{3 / 2} \cdot d t$

$\Rightarrow \frac{2 t^{\frac{3}{2}}}{3}+c$

But $\cos x=t$

$\Rightarrow \frac{2(\cos x)^{3 / 2}}{3}+c$