$\begin{aligned} & \cos ^2 x \frac{d y}{d x}-\tan 2 x \cdot y=\cos ^4 x \\ & \Rightarrow \frac{d y}{d x}-\frac{\tan 2 x}{\cos ^2 x} y=\cos ^2 x\end{aligned}$
the above differential equation is in the form of.
$$
\frac{d y}{d x}+p y=q \text {, }
$$
where $p=\frac{-\tan 2 x}{\cos ^2 x}$
and $q=\cos ^2 x$
Now, IF $=e^{\int p d x}=e^{\int-\frac{\tan 2 x}{\cos ^2 x} d x}$
Let $\begin{aligned} I & =\int-\frac{\tan 2 x}{\cos ^2 x} d x=\int \frac{-\sin 2 x}{\cos 2 x \cdot \cos ^2 x} d x \\ & =\int \frac{-2 \sin 2 x}{\cos 2 x(1+\cos 2 x)} d x\end{aligned}$
put $\cos 2 x=t$
$-2 \sin 2 x d x=d t$
$$
\begin{aligned}
I & =\int \frac{d t}{t(t+1)}=\int \frac{d t}{t^2+t+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\
& =\int \frac{d t}{\left(t+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}=\log \left|\frac{t}{t+1}\right|
\end{aligned}
$$