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.

$$
\begin{aligned}
\int_0^{2 \pi} \frac{\cos ^2 \theta-\sin ^2 \theta}{\sin ^4 \theta+\cos ^4 \theta} d \theta 
= & \int_0^{2 \pi} \frac{\cos 2 \theta}{\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-4 \sin ^2 \theta \cos ^2 \theta} d \theta \\
= & \int_0^{2 \pi} \frac{\cos 2 \theta}{1-\sin ^2 2 \theta} d \theta \\
= & \frac{1}{2} \int_0^{2 \pi} \frac{d(\sin 2 \theta)}{1-\sin ^2 2 \theta} \\
= & \frac{1}{2}\left[\ln \left|\frac{1+\sin 2 \theta}{1-\sin 2 \theta} \right|\right]_0^{2 \pi}\\
= & 0 
\end{aligned}
$$
