<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] 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.
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