Answer
$$ - \frac{{{{\cos }^6}x}}{6} + \frac{{{{\cos }^8}x}}{8} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^3}x{{\cos }^5}x} dx \cr
& {\text{split off }}\sin x \cr
& = \int {{{\sin }^2}x{{\cos }^5}x} \sin xdx \cr
& {\text{pythagorean identity}} \cr
& = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^5}x} \sin xdx \cr
& u = \cos x,{\text{ }}du = - \sin xdx \cr
& = - \int {\left( {1 - {u^2}} \right){u^5}} du \cr
& = - \int {\left( {{u^5} - {u^7}} \right)} du \cr
& {\text{evaluate the integral}} \cr
& = - \frac{{{u^6}}}{6} + \frac{{{u^8}}}{8} + C \cr
& {\text{replace }}u = \cos x \cr
& = - \frac{{{{\cos }^6}x}}{6} + \frac{{{{\cos }^8}x}}{8} + C \cr} $$