Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ How do we integrate the following?

$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$

I tried to simplify this, but I cannot seem to proceed further than the below form:

$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos2x}dx}}$
$\implies \frac{1}{2\sqrt2}\log |\sec 2x + \tan 2x| + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos^2x-\sin^2x}dx} + C$

The answer that I'm supposed to get is:

$\frac{x}{\sqrt2}+C$

Please help, thanks!

If $\cos 2x > 0$ and the integrand is wrong as Rudr Pratap Singh points out, then:
\begin{align}
\frac{\cos^4 x-\sin^4 x}{\sqrt{1+\cos 4x}}
&= \frac{(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)}{\sqrt{2\cos^2 2x}} \\
&= \frac{(1)(\cos 2x)}{\sqrt2|\cos2x|} \\
&= \frac 1 {\sqrt2}.
\end{align}
Hence,
$$\int \frac{\cos^4x-\sin^4x}{\sqrt{1+\cos 4x}} \ \text dx = \frac x {\sqrt2} +c,$$
as desired by O.P.
