Answer
$$ - \frac{{{{\cos }^3}x}}{3} + \frac{{{{\cos }^5}x}}{5} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^3}x{{\cos }^2}x} dx \cr
& {\text{split off }}\sin x \cr
& = \int {{{\sin }^2}x{{\cos }^2}x} \sin xdx \cr
& {\text{pythagorean identity}} \cr
& = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^2}x} \sin xdx \cr
& u = \cos x,{\text{ }}du = - \sin xdx \cr
& = - \int {\left( {1 - {u^2}} \right){u^2}} du \cr
& = - \int {\left( {{u^2} - {u^4}} \right)} du \cr
& {\text{evaluate the integral}} \cr
& = - \frac{{{u^3}}}{3} + \frac{{{u^5}}}{5} + C \cr
& {\text{replace }}u = \cos x \cr
& = - \frac{{{{\cos }^3}x}}{3} + \frac{{{{\cos }^5}x}}{5} + C \cr} $$
