Answer
$$\frac{{{{\sin }^5}x}}{5} - \frac{{{{\sin }^3}x}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^3}x{{\cos }^2}} xdx \cr
& {\text{split off }}{\sin ^3}x \cr
& = \int {{{\sin }^2}x{{\cos }^2}} x\sin xdx \cr
& {\text{identity }}{\sin ^2}x + {\cos ^2}x = 1 \cr
& = \int {{{\sin }^2}x{{\cos }^2}x} \sin xdx \cr
& = \int {\left( {1 - {{\cos }^2}x} \right)} co{s^2}x\sin xdx \cr
& = \int {\left( {{{\cos }^2}x - {{\cos }^4}x} \right)} \sin xdx \cr
& {\text{substitute }}u = \cos x,{\text{ }}du = - \sin xdx \cr
& = \int {\left( {{u^2} - {u^4}} \right)} \left( { - du} \right) \cr
& = \int {\left( {{u^4} - {u^2}} \right)} du \cr
& {\text{find the antiderivatives by the power rule}} \cr
& = \frac{{{u^5}}}{5} - \frac{{{u^3}}}{3} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \sin x \cr
& = \frac{{{{\sin }^5}x}}{5} - \frac{{{{\sin }^3}x}}{3} + C \cr} $$
