Answer
$$\frac{{{{\sin }^3}2x}}{6} - \frac{{{{\sin }^5}2x}}{{10}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}2x{{\cos }^3}2x} dx \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& u = \sin 2x,\,\,\,\,du = 2\cos 2xdx,\,\,\,\,dx = \frac{{du}}{{2\cos 2x}} \cr
& {\text{then}} \cr
& \int {{{\sin }^2}2x{{\cos }^3}2x} dx = \int {{u^2}{{\cos }^3}2x} \left( {\frac{{du}}{{2\cos 2x}}} \right) \cr
& = \frac{1}{2}\int {{u^2}{{\cos }^2}2x} du \cr
& {\text{use the identity co}}{{\text{s}}^2}\theta + {\sin ^2}\theta = 1 \cr
& = \frac{1}{2}\int {{u^2}\left( {1 - {{\sin }^2}2x} \right)} du \cr
& = \frac{1}{2}\int {{u^2}\left( {1 - {u^2}} \right)} du \cr
& = \frac{1}{2}\int {\left( {{u^2} - {u^4}} \right)} du \cr
& \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left( {\frac{{{u^3}}}{3} - \frac{{{u^5}}}{5}} \right) + C \cr
& = \frac{{{u^3}}}{6} - \frac{{{u^5}}}{{10}} + C \cr
& \cr
& {\text{write the integrand in terms of }}x,{\text{ replace }}\sin 2x{\text{ for }}u \cr
& = \frac{{{{\sin }^3}2x}}{6} - \frac{{{{\sin }^5}2x}}{{10}} + C \cr} $$
