Answer
$$\int \sin 2 x \cos 4 x d x =\frac{1}{4}\cos 2x -\frac{1}{12}\cos 6x+c$$
Work Step by Step
$$
\int \sin 2 x \cos 4 x d x
$$
Since
$$
\sin m x \cos n x=\frac{1}{2}(\sin [(m-n) x]+\sin [(m+n) x])
$$
Then
\begin{align*}
\int \sin 2 x \cos 4 x d x&=\frac{1}{2}\int (\sin (-2x)+\sin (6x))dx\\
&=\frac{1}{2}\int (-\sin (2x)+\sin (6x))dx\\
&=\frac{1}{2}\left( \frac{1}{2}\cos 2x -\frac{1}{6}\cos 6x\right) +c\\
&=\frac{1}{4}\cos 2x -\frac{1}{12}\cos 6x+c
\end{align*}
