<s>[INST] Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{c^2-s^2\over s^4+c^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{(1-s^2)-s^2\over s^4+(c^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(1-s^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(s^2-1)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+s^4-2s^2+1}\mathrm d\theta\\
&=\int_{0}^{2\pi}\underbrace{\color{red}{\left({1-2s^2\over 2s^4-2s^2+1}\right)}}_{\text{I got stuck here}}\mathrm d\theta\\
\end{align}$$
I need your help.
 [/INST] Graph the integrand and observe that there is "as much area above as below the x-axis".  This leads to the conjecture that the value is zero, and also gives a strategy for proving it using a symmetry argument.
First observe that the function is periodic with period $\pi$.  You can confirm this by showing that $f(x + \pi) = f(x)$ for all $x$.  Since your integral covers two periods we can say that your integral is twice the integral over $[-\frac{\pi}{4}, \frac{3\pi}{4}]$.
Then observe that the graph of $f$ is symmetric with respect to the point $(\frac{\pi}{4},0)$.  You can confirm this by showing that $f(\frac{\pi}{4} - x) = - f(\frac{\pi}{4} + x)$.
Thus the integral over $[-\frac{\pi}{4}, \frac{3\pi}{4}]$ is equal to zero.
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