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?
 A: You may just apply $t=\tan x$, $dx = \frac{dt}{1+t^2}$ and the resulting expression is quite manageable,,
$$I=\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{1-\frac12\sin^2(2x)}
= \int \frac {\frac{1-t^2}{1+t^2}\cdot\frac{dt}{1+t^2}}{1-\frac12\left(\frac{2t}{1+t^2}\right)^2}=\int \frac{1-t^2}{1+t^4}dt$$
Next, carry out the integral as follows,
$$I=-\int \frac{1 - \frac1{t^2}}{t^2+\frac1{t^2}}dt
=-\int \frac{d(t + \frac1{t})}{(t+\frac1{t})^2-2}
=\frac1 {\sqrt2} \coth^{-1} \left(\frac{t^2 + 1}{\sqrt2\>t}\right)+C $$
where $(\coth^{-1}t)' = \frac1{1-t^2}$ is used in the last step.
A: 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.
A: 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. 
A: Continues, we have $$ \int{ \frac{1}{2 - t^2} dt } $$.
assume $ t = \sqrt{2} \sin (y) $, we differensial and get $ dt = \sqrt{2} \cos (y) dy $
$$ \int{ \frac{1}{2 - t^2} dt } = \frac{1}{2} \int{ \frac{\sqrt{2} \cos (y) dy}{1 - \sin ^2 (y) } dt } = \frac{\sqrt{2}}{2} \int{ \sec(y) dy } = \frac{\sqrt{2}}{2} \sec y \tan y = \frac{1}{2- t^2}$$ . back to question
A: $\cos^4 x = \frac 18 (\cos 4x + 4\cos 2x + 3)\\
\sin^4 x = \frac 18 (\cos 4x - 4\cos 2x + 3)\\
\cos^4 x+ \sin^4 x = \frac 14(\cos 4x + 3)$
$\frac {4\cos 2x}{\cos 4x + 3}$
That looks a little better.
Lets say $\cos 4x = 1 - 2\sin^2 2x$
$\int \frac {4\cos 2x}{4 - 2\sin^2 2x}\ dx$
Now we can do a u-subtitution
$u = \sin 2x, du = 2\cos 2x$
$\int \frac {2}{4 - 2u^2}\ du$
Separate into partial fractions
$\frac 1{2\sqrt 2} \int \frac {1}{\sqrt 2-u} + \frac {1}{\sqrt 2+u} \ du$
$\frac 1{2\sqrt 2} (\ln (\sqrt 2+u)-\ln (\sqrt 2-u))$
$\frac 1{2\sqrt 2} (\ln (\sqrt 2+\sin 2x)-\ln (\sqrt 2-\sin 2x))$
