How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$? $$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$
I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows:
$\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2}x = 1 - \frac{1}{2}\cdot\sin^{2}(2x) = \frac{1 + \cos^{2}(2x)}{2}$, 
and then using the $\tan(2x) = t$ substitution. But if I do the same with the definite integral, both bounds of the integral become $0$.
 A: We have $\displaystyle\cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x=1-2\cos^2x\sin^2x$ 
$\displaystyle=\frac{2-\sin^22x}2=\frac{2(1+\tan^22x)-\tan^22x}{2\sec^22x}=\frac{\tan^22x+2}{2\sec^22x}$
$$\int\frac{dx}{\cos^4x+\sin^4x}=\int\frac{2\sec^22x}{\tan^22x+2}dx$$
Setting $\tan2x=u,$ $$\int\frac{2\sec^22x}{\tan^22x+2}dx=\int\frac{du}{u^2+(\sqrt2)^2}=\frac1{\sqrt2}\arctan\left(\frac u{\sqrt2}\right)+K$$
$$\implies \int\frac{dx}{\cos^4x+\sin^4x}=\frac1{\sqrt2}\arctan\left(\frac{\tan2x}{\sqrt2}\right)+K\  \ \ \  (1)$$
Now $\displaystyle\tan2x=0\iff 2x=n\pi\iff x=\frac{n\pi}2$ where $n$ is any integer
Establish that $$\int_0^{2a}f(x)dx=\begin{cases} 2\int_0^af(x)dx &\mbox{if } f(2a-x)=f(x) \\ 
0 & \mbox{if } f(2a-x)=-f(x) \end{cases} $$
Setting $2a=2\pi\iff a=\pi$  and $\displaystyle f(x)=\cos^4x+\sin^4x$
$\displaystyle\cos(2\pi-x)=\cos x,\sin(2\pi-x)=-\sin x\implies f(2\pi-x)=f(x)$
$$\implies I=\int_0^{2\pi}\frac{dx}{\cos^4x+\sin^4x}=2\int_0^{\pi}\frac{dx}{\cos^4x+\sin^4x}$$
Again setting $\displaystyle2a=\pi\iff a=\frac\pi2$
$\displaystyle\cos(\pi-x)=-\cos x,\sin(\pi-x)=+\sin x\implies f(\pi-x)=f(x)$
$$\implies I=2\int_0^{\pi}\frac{dx}{\cos^4x+\sin^4x}2=2\cdot2\int_0^{\dfrac\pi2}\frac{dx}{\cos^4x+\sin^4x}$$
Finally set $\displaystyle2a=\frac\pi2\iff a=\frac\pi4$
$\displaystyle\implies\cos\left(\frac\pi2-x\right)=\sin x,\sin\left(\frac\pi2-x\right)=\cos x$
$\displaystyle\implies f(x)=f\left(\frac\pi2-x\right)$
$\displaystyle\implies I=4\cdot2\int_0^{\dfrac\pi4}\frac{dx}{\cos^4x+\sin^4x}$
From $\displaystyle(1),I=8\left[\frac1{\sqrt2}\arctan\left(\frac{\tan2x}{\sqrt2}\right)+K\right]_0^{\frac\pi4}=\frac8{\sqrt2}\left(\frac\pi2-0\right)$
A: Note that $\tan 2x$ is discontinuous at $\frac{\pi}4$ and some other values which can be easily found.
You have to break the integral from $0$ to $\frac{\pi}4$ and so on.
A: Let $z=e^{i x}$; then $dx = -i dz/z$ and the integral is equal to
$$-i 8 \oint_{|z|=1} dz \frac{z^3}{z^8 + 6 z^4+1}$$
By the residue theorem, the integral is then equal to $i 2 \pi$ times the sum of the residues at each pole inside the unit circle.  The residue at each pole $z_k$ is equal to
$$-i 8 \frac{z_k^3}{8 z_k^7 + 24 z_k^3} = -i \frac1{z_k^4+3}$$
Each pole $z_k$ inside the unit circle satisfies $z_k^4+3=2 \sqrt{2}$, so the integral is therefore
$$2 \pi \frac{4}{2 \sqrt{2}} = 2 \sqrt{2} \pi$$ 
A: \begin{aligned}
& \int_{0}^{2 \pi} \frac{d x}{\sin ^{4} x+\cos ^{4} x} \\
=& \int_{0}^{2 \pi} \frac{d x}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x} \\
=& \int_{0}^{2 \pi} \frac{d x}{1-\frac{\sin ^{2} 2 x}{2}} \\
=& 16 \int_{0}^{\frac{\pi}{4}} \frac{d x}{1+\cos ^{2} 2 x} \\
=& 16 \int_{0}^{\frac{\pi}{4}} \frac{\sec ^{2} 2 x}{\sec ^{2} 2 x+1}d x \\
=& 8 \int_{0}^{\frac{\pi}{4}} \frac{d(\tan 2 x)}{\tan ^{2}(2 x)+2}
\\=&4 \sqrt{2}\left[\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)\right]_{0}^{\frac{\pi}{4}} \\
=& 2 \sqrt{2} \pi
\end{aligned}
