Finding $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$ $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$
I'm looking more into simplifying this than the solution itself (which I know involves using t=tan(x/2)).
I did:
$$\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x} = \int\frac{\cos(2x)dx}{1-2\sin^2(2x)}$$
I tried applying the t=tan(y/2) (where y = 2x) but it became a mess so I figure this can be simplified further... help?

You are almost there. Set $t=\sin(2x)$ so that $\frac12dt=\cos(2x)dx$. Then, as @Azif00 noted, you integral would be $$\int\frac{1}{1-\frac12 t^2}\frac{dt}2=\int\frac{dt}{2-t^2}.$$
You can compute this with partial fractions or a substitution.
Alternatively, you could use $u=\tan(x)$ so that 
$$\sin(2x)=\frac{2u}{1+u^2},$$
$$\cos(2x)=\frac{1-u^2}{1+u^2},$$
and $$dx=\frac{2du}{1+u^2}.$$
Then we have your integral as
$$2\int\frac{1}{1-\frac12(\tfrac{2u}{1+u^2})^2}\frac{1-u^2}{(1+u^2)^2}du\\
=2\int\frac{1-u^2}{u^4+1}du,$$
the rest of which I leave to you as an exercise. 
