$$
\begin{aligned}
& \text { Let } I=\int \frac{\cos 2 x \sin 4 x}{\cos ^4 x\left(1+\cos ^2 2 x\right)} d x \\
& =4 \int \frac{\cos 2 x(2 \sin 2 x \cos 2 x)}{(1+\cos 2 x)^2\left(1+\cos ^2 2 x\right)} d x
\end{aligned}
$$
Put $\cos 2 x=t \Rightarrow-2 \sin 2 x d x=d t$
$$
\begin{aligned}
\therefore I & =-4 \int \frac{t^2}{(1+t)^2\left(1+t^2\right)} d t \\
& =-4 \int\left(\frac{1}{2(1+t)^2}-\frac{1}{2(1+t)}+\frac{t}{2\left(1+t^2\right)}\right) d t \\
& =-2\left[\frac{-1}{1+t}-\log (1+t)+\frac{1}{2} \operatorname{Iog}\left(1+t^2\right)\right]+c \\
& =\frac{2}{1+\cos 2 x}+2 \log (1+\cos 2 x) \\
& =\frac{2}{2 \cos ^2 x}+\log \frac{\left(1+\cos ^2 x\right)^2}{\left(1+\cos ^2 2 x\right)}+c \\
& =\log \left(\frac{\left(1+\cos ^2 x\right)^2}{\left(1+\cos ^2 2 x\right)}\right)+\sec ^2 x+c
\end{aligned}
$$