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!
 A: Use $$\frac{\sin^4x+\cos^4x}{\sqrt{1+\cos4x}}=\frac{1-\frac{1}{2}\sin^22x}{\sqrt2\cos2x}=\frac{\frac{1}{2}+\frac{1}{2}\cos^22x}{\sqrt2\cos2x}=\frac{\cos2x}{2\sqrt2(1-\sin^22x)}+\frac{1}{2\sqrt2}\cos2x=$$
$$=\frac{1}{4\sqrt2}\left(\frac{\cos2x}{1+\sin2x}+\frac{\cos2x}{1-\sin2x}\right)+\frac{1}{2\sqrt2}\cos2x.$$
A: 
$$\int { \frac { \cos ^{ 4 } x+\sin ^{ 4 } x }{ \sqrt { 1+\cos4x }  } dx } =\frac { 1 }{ \sqrt { 2 }  } \int { \frac { { \left( \sin ^{ 2 }{ x+\cos ^{ 2 }{ x }  }  \right)  }^{ 2 }-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x }  }  }{ \sqrt { \cos ^{ 2 }{ 2x }  }  } dx } =\\ =\frac { 1 }{ \sqrt { 2 }  } \int { \frac { 1-\frac { \sin ^{ 2 }{ 2x }  }{ 2 }  }{ \cos { 2x }  }  } dx=\frac { 1 }{ 2\sqrt { 2 }  } \int { \frac { 2-\sin ^{ 2 }{ 2x }  }{ \cos { 2x }  }  } dx=\\ =\frac { 1 }{ 2\sqrt { 2 }  } \int { \frac { \cos ^{ 2 }{ 2x } +1 }{ \cos { 2x }  }  } dx=\frac { 1 }{ 2\sqrt { 2 }  } \left[ \int { \cos { 2xdx }  } +\int { \frac { dx }{ \cos { 2x }  }  }  \right] =\\ =\frac { 1 }{ 4\sqrt { 2 }  } \int { d\left( \sin { 2x }  \right)  } +\frac { 1 }{ 2\sqrt { 2 }  } \int { \frac { \cos { 2xdx }  }{ \cos ^{ 2 }{ 2x }  }  } =\\ =\frac { \sin { 2x }  }{ 4\sqrt { 2 }  } +\frac { 1 }{ 4\sqrt { 2 }  } \int { \frac { d\left( \sin { 2x }  \right)  }{ 1-\sin ^{ 2 }{ 2x }  }  } =\frac { \sin { 2x }  }{ 4\sqrt { 2 }  } +\frac { 1 }{ 8\sqrt { 2 }  } \left[ \int { \frac { d\left( \sin { 2x }  \right)  }{ 1-\sin { 2x }  }  } +\int { \frac { d\left( \sin { 2x }  \right)  }{ 1+\sin { 2x }  }  }  \right] =\\ =\frac { \sin { 2x }  }{ 4\sqrt { 2 }  } +\frac { 1 }{ 8\sqrt { 2 }  } \left[ \int { \frac { d\left( 1+\sin { 2x }  \right)  }{ 1+\sin { 2x }  }  } -\int { \frac { d\left( 1-\sin { 2x }  \right)  }{ 1-\sin { 2x }  }  }  \right] =\\ =\frac { \sin { 2x }  }{ 4\sqrt { 2 }  } +\frac { 1 }{ 8\sqrt { 2 }  } \ln { \left| \frac { 1+\sin { 2x }  }{ 1-\sin { 2x }  }  \right| + } C\\ $$

A: Using $\cos2x=1-2\sin^2x=2\cos^2x-1,$
$$I=\dfrac{\sin^4x+\cos^4x}{\sqrt{1+\cos4x}}=\dfrac{(1-\cos2x)^2+(1+\cos2x)^2}{4\sqrt2|\cos2x|}=\dfrac{1+\cos^22x}{2\sqrt2|\cos2x|}$$
For $\cos2x>0,$
$$2\sqrt2I=\sec2x+\cos2x$$
Now use Integral of the secant function
A: I figured out why nobody is getting that answer: the question is wrong, the (correct) question is
$$\int \frac{\cos^4 x-\sin^4 x}{\sqrt{1+\cos4x}} \ \text{d}x$$
A: 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.
