Answer
$$\frac{1}{24}$$
Work Step by Step
Given
$$ \int_{0}^{\pi/6}\sin 2 x\cos 4xdx$$
Use
$$ \sin (a x) \cos (b x)=\frac{1}{2} \sin ((a-b) x)+\frac{1}{2} \sin ((a+b) x) $$
Then
\begin{align*}
\int_{0}^{\pi/6}\sin 2 x\cos 4xdx&=\frac{1}{2}\int_{0}^{\pi}(\sin (- 2x)+ \sin 6x)dx\\
&=\frac{1}{2}\int_{0}^{\pi}( \sin 6x-\sin 2x)dx\\
&=\frac{ -1}{2}\left(\frac{1}{6}\cos6 x-\frac{1}{2}\cos2 x\right)\bigg|_{0}^{\pi} \\
&=\frac{1}{24}
\end{align*}
