Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$ 
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$

My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\pi}\frac {1}{\cos(\frac{x}2)}dx=\int_0^{2\pi}\frac 1{\cos x}dx=0$
So it actually is:
$$I=2\int_0^{2\pi}\frac {1}{2-\sin^2(2x)}dx=\int_0^{4\pi}\frac{1}{2-\sin^2(x)}dx=\int_0^{4\pi}\frac 1{1+\cos^2x}dx$$
Now if I try to make the substituion $u=\tan(\frac x2)$ I get integral from $0$ to $0$...Why?
What I am doing wrong?

One easy approach to this problem is by dividing both numerator and denominator by $\dfrac{\tan^4x}{\tan^4x}$
$$\dfrac{1}{\sin^4x+\cos^4x}\left(\dfrac{\tan^4x}{\tan^4x}\right)=\dfrac{\sec^4x}{1+\tan^4x}=\dfrac{(1+\tan^2x)\sec^2x}{1+\tan^4x}$$ 
Now you can use $u-$ substitution $u=\tan x$
Can you take it from here?
