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?

We know that
$$\begin{align}
\cos(2x) &= \cos^2(x) + \sin^2(x)\\
\sin(x) &= \dfrac{\tan(x)}{\sec(x)}\\
\sec^2(x) &= 1 + \tan^2(x)
\end{align}$$
So,
$$\int\dfrac{\cos(2x)}{\sin^4(x)+\cos^4(x)}\,\mathrm dx \equiv \int\sec^2x\left(\dfrac{-(\tan(x) - 1)(\tan(x) + 1)}{\tan^4(x) + 1}\right)\,\mathrm dx$$
Let $u = \tan(x)$. So, $\dfrac{\mathrm du}{\mathrm dx} = \sec^2(x)\to\mathrm dx = \dfrac{\mathrm du}{\sec^2(x)}$.
$$\implies\int\sec^2x\left(\dfrac{-(\tan(x) - 1)(\tan(x) + 1)}{\tan^4(x) + 1}\right)\,\mathrm dx\equiv-\int\dfrac{(u - 1)(u + 1)}{u^4 + 1}\mathrm du$$
Now, you should be able to factor the denominator and use partial fractions to get the final answer.
