Answer
$$\frac{7}{{24}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /6} {\sin 4x\cos 2x} dx \cr
& {\text{product }}\sin mx\cos nx = \frac{1}{2}\left( {\sin \left( {\left( {m - n} \right)x} \right) + \sin \left( {\left( {m + n} \right)x} \right)} \right) \cr
& \sin 4x\cos 2x = \frac{1}{2}\left( {\sin \left( {\left( {4 - 2} \right)x} \right) + \sin \left( {\left( {4 + 2} \right)x} \right)} \right) \cr
& \sin 4x\cos 2x = \frac{1}{2}\left( {\sin 2x + \sin 6x} \right) \cr
& \int_0^{\pi /6} {\sin 4x\cos 2x} dx = \frac{1}{2}\int_0^{\pi /6} {\left( {\sin 2x + \sin 6x} \right)} dx \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{2}\left[ { - \frac{{\cos 2x}}{2} - \frac{{\sin 6x}}{6}} \right]_0^{\pi /6} \cr
& {\text{fundamental theorem of calculus}} \cr
& = \frac{1}{2}\left[ { - \frac{{\cos 2\left( {\pi /6} \right)}}{2} - \frac{{\cos 6\left( {\pi /6} \right)}}{6}} \right] - \frac{1}{2}\left[ { - \frac{{\cos 2\left( {\pi /6} \right)}}{2} - \frac{{\cos 6\left( {\pi /6} \right)}}{6}} \right] \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\left[ { - \frac{{1/2}}{2} - \frac{{ - 1}}{6}} \right] - \frac{1}{2}\left[ { - \frac{1}{2} - \frac{1}{6}} \right] \cr
& = \frac{1}{2}\left[ { - \frac{1}{{12}}} \right] - \frac{1}{2}\left[ { - \frac{2}{3}} \right] \cr
& = - \frac{1}{{24}} + \frac{1}{3} \cr
& = \frac{7}{{24}} \cr} $$
