Answer
$$\frac{1}{{24}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /6} {\sin 2x} \cos 4xdx \cr
& {\text{product }}\sin mx\cos nx = \frac{1}{2}\left( {\sin \left( {\left( {m - n} \right)x} \right) + \sin \left( {\left( {m + n} \right)x} \right)} \right) \cr
& \sin 2x\cos 4x = \frac{1}{2}\left( {\sin \left( {\left( {2 - 4} \right)x} \right) + \sin \left( {\left( {2 + 4} \right)x} \right)} \right) \cr
& \sin 2x\cos 4x = \frac{1}{2}\left( { - \sin 2x + \sin 6x} \right) \cr
& \int {\sin x\cos 4x} dx = \frac{1}{2}\int {\left( {\sin 6x - \sin 2x} \right)} dx \cr
& {\text{sum rule}} \cr
& = \frac{1}{2}\int {\sin 6x} dx - \frac{1}{2}\int {\sin 2x} dx \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{2}\left( { - \frac{1}{6}\cos 6x} \right) - \frac{1}{2}\left( { - \frac{1}{2}\cos 2x} \right) + C \cr
& = - \frac{1}{{12}}\cos 6x + \frac{1}{4}\cos 2x + C \cr
& \int_0^{\pi /6} {\sin 2x} \cos 4xdx = \left[ { - \frac{1}{{12}}\cos 6x + \frac{1}{4}\cos 2x} \right]_0^{\pi /6} \cr
& {\text{fundamental theorem}} \cr
& = \left[ { - \frac{1}{{12}}\cos 6\left( {\frac{\pi }{6}} \right) + \frac{1}{4}\cos 2\left( {\frac{\pi }{6}} \right)} \right] - \left[ { - \frac{1}{{12}}\cos \left( 0 \right) + \frac{1}{4}\cos \left( 0 \right)} \right] \cr
& = \left[ {\frac{1}{{12}} + \frac{1}{4}\left( {\frac{1}{2}} \right)} \right] - \left[ { - \frac{1}{{12}} + \frac{1}{4}} \right] \cr
& = \frac{5}{{24}} - \frac{1}{6} \cr
& = \frac{1}{{24}} \cr} $$
