Answer
$$\frac{3}{8}x - \frac{1}{8}\sin 4x + \frac{{\sin 8x}}{{64}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^4}2x} dx \cr
& = \int {{{\left( {{{\sin }^2}2x} \right)}^2}} dx \cr
& {\text{trigonometric identity si}}{{\text{n}}^2}\theta = \frac{{1 - \cos 2\theta }}{2} \cr
& = \int {{{\left( {\frac{{1 - \cos 4x}}{2}} \right)}^2}} dx \cr
& = \frac{1}{4}\int {{{\left( {1 - \cos 4x} \right)}^2}} dx \cr
& = \frac{1}{4}\int {\left( {1 - 2\cos 4x + {{\cos }^2}4x} \right)} dx \cr
& {\text{trigonometric identity co}}{{\text{s}}^2}\theta = \frac{{1 + \cos 2\theta }}{2} \cr
& = \frac{1}{4}\int {\left( {1 - 2\cos 4x + \frac{{1 + \cos 8x}}{2}} \right)} dx \cr
& = \frac{1}{4}\int {\left( {1 - 2\cos 4x + \frac{1}{2} + \frac{{\cos 8x}}{2}} \right)} dx \cr
& = \frac{1}{4}\int {\left( {\frac{3}{2} - 2\cos 4x + \frac{{\cos 8x}}{2}} \right)} dx \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{4}\left( {\frac{3}{2}x - \frac{1}{2}\sin 4x + \frac{{\sin 8x}}{{16}}} \right) + C \cr
& = \frac{3}{8}x - \frac{1}{8}\sin 4x + \frac{{\sin 8x}}{{64}} + C \cr} $$
