<s>[INST] 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?
 [/INST] $\cos^4 x = \frac 18 (\cos 4x + 4\cos 2x + 3)\\
\sin^4 x = \frac 18 (\cos 4x - 4\cos 2x + 3)\\
\cos^4 x+ \sin^4 x = \frac 14(\cos 4x + 3)$
$\frac {4\cos 2x}{\cos 4x + 3}$
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
$\frac 1{2\sqrt 2} \int \frac {1}{\sqrt 2-u} + \frac {1}{\sqrt 2+u} \ du$
$\frac 1{2\sqrt 2} (\ln (\sqrt 2+u)-\ln (\sqrt 2-u))$
$\frac 1{2\sqrt 2} (\ln (\sqrt 2+\sin 2x)-\ln (\sqrt 2-\sin 2x))$
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