Evaluate the following integrals:

Let $\cos x=t$ then $d t=-\sin x d x$

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

Substitute all these in the above equation,

$\int \sin ^{3} x \cos ^{5} x d x=\int \sin ^{3} x t^{5}\left(-\frac{d t}{\sin x}\right)$

$=-\int \sin ^{2} \mathrm{xt}^{5} \mathrm{dt}$

$=-\int\left(1-\cos ^{2} \mathrm{x}\right) \mathrm{t}^{5} \mathrm{dt}$

$=-\int\left(1-\mathrm{t}^{2}\right) \mathrm{t}^{5} \mathrm{dt}$

$=-\int \mathrm{t}^{5} \mathrm{~d} \mathrm{t}-\int \mathrm{t}^{7} \mathrm{dt}$

$=-\frac{t^{6}}{6}+\frac{t^{8}}{8}+c\left(\right.$ since $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$ for any $\left.c \neq-1\right)$

$=-\frac{\cos ^{6} x}{6}+\frac{\cos ^{8} x}{8}+c$

$=\frac{1}{8} \cos ^{8} x-\frac{1}{6} \cos ^{6} x+c$