Evaluate the following integrals:

Assume $\sin x=t$

$d(\sin x)=d t$

$\cos x d x=d t$

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

$\Rightarrow \int \mathrm{t}^{5} \mathrm{dt}$

$\Rightarrow \frac{t^{6}}{6}+c$

But $t=\sin x$

$\Rightarrow \frac{\sin ^{6} x}{6}+c$