Integrate arctan from $0$ to $3\pi$ How do I integrate
$$\int_0^{3\pi} \frac 1 {\sin^4x + \cos^4 x} \,dx$$ 
I tried with Weierstrass and obtained:$$
\int\frac 1 {u^2+2}\, du
$$ I think it's correct but how do I integrate this given that I cant integrate arctan for $3\pi$ and $0$

$\sin^4(x)+\cos^4(x)=(\sin^2(x)+\cos^2(x))^2-2\sin^2(x)\cos^2(x) = 1-\frac{1}{2}\sin^2(2x) $ leads to:
$$ I = \frac{1}{2}\int_{0}^{6\pi}\frac{dz}{1-\frac{1}{2}\sin^2(z)}=6\int_{0}^{\pi/2}\frac{dz}{1-\frac{1}{2}\sin^2(z)}=6\int_{0}^{\pi/2}\frac{dz}{1-\frac{1}{2}\cos^2(z)} $$
by periodicity and symmetry. By setting $z=\arctan t$ the last integral turns into:
$$ I = 6 \int_{0}^{+\infty}\frac{dt}{\frac{1}{2}+t^2} = \color{red}{3\pi\sqrt{2}}.$$
