Evaluate the following integrals:

Let $\sin x=t$

Then $d(\sin x)=d t=\cos x d x$

Put $t=\sin x$ and $d t=\cos x d x$ in above equation

$\int \sin ^{5} x \cos x d x=\int t^{5} d t$

$=\frac{t^{6}}{6}+c\left(\right.$ since $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$ for any $\left.c \neq-1\right)$

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