Answer
$$ - \frac{1}{{12}}\cos 6x + \frac{1}{4}\cos 2x + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 2x} \cos 4xdx \cr
& {\text{using the special product }}\sin Ax\cos Bx = \frac{1}{2}\left[ {\sin \left( {A + B} \right)x + \sin \left( {A - B} \right)x} \right] \cr
& \int {\sin 2x} \cos 4xdx = \int {\frac{1}{2}\left[ {\sin \left( {2 + 4} \right)x + \sin \left( {2 - 4} \right)x} \right]} dx \cr
& = \frac{1}{2}\int {\left( {\sin 6x - \sin 2x} \right)} dx \cr
& = \frac{1}{2}\int {\sin 6x} dx - \frac{1}{2}\int {\sin 2x} dx \cr
& \cr
& {\text{integrate by the formula }}\int {\sin ax} dx = - \frac{1}{a}\cos ax + C \cr
& = \frac{1}{2}\left( { - \frac{1}{6}\cos 6x} \right) - \frac{1}{2}\left( { - \frac{1}{2}\cos 2x} \right) + C \cr
& \cr
& {\text{simplifying, we get}} \cr
& = - \frac{1}{{12}}\cos 6x + \frac{1}{4}\cos 2x + C \cr} $$
