Solution:

Let $\cos x=t$
$\begin{array}{l}
\Rightarrow \mathrm{d}(\cos x)=d t \\
\Rightarrow-\sin x d x=d t \\
\Rightarrow d x=\frac{-d t}{\sin x}
\end{array}$
$\therefore$ On substituting $\mathrm{t}$ and $\mathrm{dt}$ in the given equation we obtain
$\Rightarrow \int \sqrt[3]{t^{2}} \sin x \cdot \frac{d t}{\sin x}$
$\begin{array}{l}
\Rightarrow \int-\mathrm{t}^{2 / 3} \cdot \mathrm{dt} \\
\Rightarrow-\frac{3}{5} t^{\frac{5}{3}}+c
\end{array}$
But $\cos x=t$
$\Rightarrow-\frac{3}{5} \cos ^{\frac{5}{3}} x+c$