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 $0$.

\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}
