Answer
$$\frac{{{{\cos }^5}2x}}{{10}} - \frac{{{{\cos }^3}2x}}{6} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^3}2x{{\cos }^2}2x} dx \cr
& {\text{split off }}{\sin ^3}2x \cr
& = \int {{{\sin }^2}2x{{\cos }^2}} 2x\sin 2xdx \cr
& {\text{identity }}{\sin ^2}2x + {\cos ^2}2x = 1 \cr
& = \int {{{\sin }^2}2x{{\cos }^2}2x} \sin 2xdx \cr
& = \int {\left( {1 - {{\cos }^2}2x} \right)} co{s^2}2x\sin 2xdx \cr
& = \int {\left( {{{\cos }^2}2x - {{\cos }^4}2x} \right)} \sin 2xdx \cr
& {\text{substitute }}u = \cos 2x,{\text{ }}du = - 2\sin xdx \cr
& = \int {\left( {{u^2} - {u^4}} \right)} \left( { - \frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int {\left( {{u^4} - {u^2}} \right)} du \cr
& {\text{find the antiderivatives by the power rule}} \cr
& = \frac{{{u^5}}}{{10}} - \frac{{{u^3}}}{6} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \cos 2x \cr
& = \frac{{{{\cos }^5}2x}}{{10}} - \frac{{{{\cos }^3}2x}}{6} + C \cr} $$