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.
 A: Method 1
Note that by defining $u=\theta-\dfrac{\pi}{2}$, we have
$$
\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
=
\int_{\dfrac{\pi}{2}}^{2\pi+\dfrac{\pi}{2}}{\sin^2u-\cos^2u\over\cos^4 u+\sin^4 u}\mathrm du.
$$
Now, since the integrand is periodic with a period of $2\pi$, we obtain:
$$
\int_{\dfrac{\pi}{2}}^{2\pi+\dfrac{\pi}{2}}{\sin^2u-\cos^2u\over\cos^4 u+\sin^4 u}\mathrm du
=\int_{0}^{2\pi}{\sin^2u-\cos^2u\over\cos^4 u+\sin^4 u}\mathrm du=-I,
$$
hence, $I=-I=0$.
Method 2
The indefinite integral can be found as follows:
$$
\int{\cos^2\theta-\sin^2\theta\over\sin^4\theta+\cos^4\theta} d\theta
{=
\int{\cos2\theta\over\sin^4\theta+\cos^4\theta+2\sin^2\theta\cos^2\theta-2\sin^2\theta\cos^2\theta} d\theta
\\=
\int{\cos2\theta\over(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta}d\theta
\\=\int \dfrac{\cos2\theta}{1-\dfrac{1}{2}\sin^22\theta}d\theta
\\=\int \dfrac{2\cos2\theta}{2-\sin^22\theta}d\theta
}
$$
where by considering $u=\sin 2\theta$ we achieve
$$
\int\dfrac{du}{2-u^2}{=
\int\dfrac{\dfrac{1}{2\sqrt 2}}{\sqrt2-u}+\dfrac{\dfrac{1}{2\sqrt 2}}{\sqrt2+u}du
\\=\dfrac{1}{2\sqrt 2}\ln\left|\dfrac{\sqrt 2+u}{\sqrt 2-u}\right|+C
\\=\dfrac{1}{2\sqrt 2}\ln\dfrac{\sqrt 2+\sin 2\theta}{\sqrt 2-\sin 2\theta}+C.
}
$$
Now, the value of the definite integral becomes $0$.
A: $$I=\int_0^{2\pi}\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Substitute $\theta→\pi-\theta$.
$$I=\int_{-\pi}^\pi\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Since the integrand is even, we can use even-odd properties.
$$I=2\int_{0}^\pi\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Substitute $\theta→\frac{\pi}{2}-\theta$
$$I=2\int_{-\pi/2}^{\pi/2}\frac{\sin^2\theta-\cos^2\theta}{\cos^4\theta+\sin^4\theta}$$
Knowing the integrand is even, we can use even odd properties again
$$I=4\int_{0}^{\pi/2}\frac{\sin^2\theta-\cos^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Substitute $\theta→\frac{\pi}{2}-\theta$ again
$$I=4\int_{0}^{\pi/2}\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Add two versions of I
$$I=4\int_{0}^{\pi/2}\frac{\sin^2\theta-\cos^2\theta}{\cos^4\theta+\sin^4\theta}d\theta+4\int_{0}^{\pi/2}\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
We find the 2 integrals are negatives of each other, so the whole RHS cancels out, leaving zero. We achieve
$$I=0$$
A: 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.
A: Here is an alternate solution using residues. Starting from where you left off, we can use the complex definitions of $\sin(x)$ and $\cos(x)$ to transform the integral into
$$\int_{0}^{2\pi}\frac{4e^{2ix}\left(1+e^{4ix}\right)}{6e^{4ix}+e^{8ix}+1}dx.$$
Since the integrand is periodic on $\pi$, let $z=e^{2ix}$. The contour $\gamma$ will be the unit circle. Then
$$\oint_{\gamma} \frac{4z\left(1+z^{2}\right)}{6z^{2}+z^{4}+1}\left(\frac{dz}{2iz}\right) = -2i\oint_{\gamma}\frac{1+z^{2}}{6z^{2}+z^{4}+1}dz.$$
The set of singularities on the disc of $\gamma$ is $Z := \left\{i\sqrt{3-2\sqrt{2}}, -i\sqrt{3-2\sqrt{2}}\right\}.$ By Residue Theory, we transform the contour integral into
$$2\pi i (-2i)\sum_{z_0 \in Z}\operatorname{Res}\left(\frac{1+z^{2}}{6z^{2}+z^{4}+1}, z=z_0\right).$$
Thus,
$$2\pi i\left(-2i\right)\left(\frac{1+\left(i\sqrt{3-2\sqrt{2}}\right)^{2}}{\frac{d}{dz}\left[6z^{2}+z^{4}+1\right]_{z=i\sqrt{3-2\sqrt{2}}}}\right) + 2\pi i\left(-2i\right)\left(\frac{1+\left(-i\sqrt{3-2\sqrt{2}}\right)^{2}}{\frac{d}{dz}\left[6z^{2}+z^{4}+1\right]_{z=-i\sqrt{3-2\sqrt{2}}}}\right)$$
which equals
$$2\pi i\left(-2i\right)\left(\frac{1+\left(i\sqrt{3-2\sqrt{2}}\right)^{2}}{12\left(i\sqrt{3-2\sqrt{2}}\right)+4\left(i\sqrt{3-2\sqrt{2}}\right)^{3}}\right)+2\pi i\left(-2i\right)\left(\frac{1+\left(-i\sqrt{3-2\sqrt{2}}\right)^{2}}{12\left(-i\sqrt{3-2\sqrt{2}}\right)+4\left(-i\sqrt{3-2\sqrt{2}}\right)^{3}}\right).$$
This simplifies down to $0$.
A: $$
\begin{aligned}
\int_0^{2 \pi} \frac{\cos ^2 \theta-\sin ^2 \theta}{\sin ^4 \theta+\cos ^4 \theta} d \theta 
= & \int_0^{2 \pi} \frac{\cos 2 \theta}{\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-4 \sin ^2 \theta \cos ^2 \theta} d \theta \\
= & \int_0^{2 \pi} \frac{\cos 2 \theta}{1-\sin ^2 2 \theta} d \theta \\
= & \frac{1}{2} \int_0^{2 \pi} \frac{d(\sin 2 \theta)}{1-\sin ^2 2 \theta} \\
= & \frac{1}{2}\left[\ln \left|\frac{1+\sin 2 \theta}{1-\sin 2 \theta} \right|\right]_0^{2 \pi}\\
= & 0 
\end{aligned}
$$
A: For the antiderivative, a little bit faster could be $${\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}=\frac{4 \cos (2 \theta )}{3+\cos (4 \theta )}$$
$$\theta=\tan ^{-1}(t)\quad \implies I= \int \frac{4 \cos (2 \theta )}{3+\cos (4 \theta )}\,d\theta=\int \frac{1-t^2}{t^4+1}\,dt$$
$$\frac{1-t^2}{t^4+1}=\frac{1-t^2}{(t^2+i)(t^2-i)}=-\frac{\frac{1-i}{2}}{t^2+i}-\frac{\frac{1+i}{2}}{t^2-i}$$ making
$$I=\frac{1}{2 \sqrt{2}}\log \left(\left| -\frac{t^2+\sqrt{2} t+1}{t^2-\sqrt{2}   t+1}\right| \right)$$
