How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$?

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

$$2 \pi \frac{4}{2 \sqrt{2}} = 2 \sqrt{2} \pi$$

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 $\displaystyle(1),I=8\left[\frac1{\sqrt2}\arctan\left(\frac{\tan2x}{\sqrt2}\right)+K\right]_0^{\frac\pi4}=\frac8{\sqrt2}\left(\frac\pi2-0\right)$

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.