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 $0$
 A: $\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:
$$ I = 6 \int_{0}^{+\infty}\frac{dt}{\frac{1}{2}+t^2} = \color{red}{3\pi\sqrt{2}}.$$
A: You're going from $0$ to $3\pi.$ That means going one-and-a-half times around the circle. The substitution
\begin{align}
\cos t & = \frac{1-t^2}{1+t^2} \\[8pt]
\sin t & = \frac{2t}{1+t^2}
\end{align}
goes once around the circle as $t$ goes from $-\infty$ to $+\infty.$ So one way to handle this integral is by doing that and then adding the same integral from $0$ to $+\infty$, since that covers the upper half of the circle, just as $0\le\theta\le\pi$ covers the upper half of the circle.
Another way is to argue from symmetry and trigonometric identities and the fact that the powers of sine and cosine are even numbers, to the conclusion that the integral from $0$ to $3\pi$ is just $3$ times the integral from $0$ to $\pi$, so the substitution gives you $3$ times the integral over $0\le t\le+\infty.$
