Answer
$${\text{FALSE}}$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}2x\cos 2x} dx \cr
& {\text{Let }}u = \sin 2x,{\text{ }}du = 2\cos 2xdx,{\text{ }}dx = \frac{1}{{2\cos 2x}}du \cr
& {\text{Substituting}} \cr
& \int {{{\sin }^2}2x\cos 2x} dx = \int {{u^2}\cos 2x\left( {\frac{1}{{2\cos 2x}}} \right)} du \cr
& = \int {{u^2}\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int {{u^2}} du \cr
& {\text{Integrating}} \cr
& {\text{ = }}\frac{1}{2}\left( {\frac{{{u^3}}}{3}} \right) + C \cr
& = \frac{1}{6}{u^3} + C \cr
& {\text{Write in terms of }}x,{\text{ let }}u = \sin 2x \cr
& = \frac{1}{6}{\sin ^3}2x + C \cr
& {\text{Therefore, the statement is FALSE}} \cr} $$
