Indefinite integral $ \int\frac{ \cos 2x}{\sin^4 x+\cos^4 x}\,dx$ Can someone help me with this integral:
$$\int\frac{\cos 2x}{\sin^4 x+\cos^4 x}\,dx$$ 
I don't understand why $\sin^4 x+\cos^4 x$ is not equal to $1$.
Unfortunately, Wolfram doesn't show step by step solution. 
Thank you for your help. 
 A: The reason why is that $$\left(\sin^2(x)+\cos^2(x)\right)^2=\sin^4(x)+\cos^4(x)+2\sin^2(x)\cos^2(x)=1^2=1$$ So, $$\sin^4(x)+\cos^4(x)=1-2\sin^2(x)\cos^2(x)=1-\frac 12 \sin^2(2x)$$ So $$I=\int \frac{\cos (2x)}{\sin^4 (x)+\cos^4 (x)}dx=\int \frac{\cos (2x)}{1-\frac 12 \sin^2(2x)}dx$$
Now, there is a beautiful thing here. I let you the pleasure of finding it and finishing the problem.
I am sure that you can take it from here.
A: Here are two substitutions that fit this integral well:
1) Since
$$
\frac{\cos 2x}{\cos^4x+\sin^4x}=\frac{1}{\cos^2x}\frac{1-\tan^2x}{1+\tan^4x}
$$
we let $u=\tan x$. This will transform the integral into
$$
\int \frac{1-u^2}{1+u^4}\,du.
$$
2) Next, in the spirit of @juantheron, let $v=u+1/u$. This will transform the integral into
$$
\int \frac{1}{2-v^2}\,dv=\frac{1}{\sqrt{2}}\text{artanh}\frac{v}{\sqrt{2}}+C.
$$
I leave it to you to do the substitutions back to the $x$ variable.
A: HINT:
$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\dfrac{(\sin2x)^2}2$$
Let $\sin2x=u$
