Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $ 
Prove that $$\int_0^\frac{n\pi}{4} \left(\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}\right) dx = \frac{n \pi}{4} $$
  is true for all integers $n$.

Here I've looked at the case where $n=1$ as $t=\tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.
$$
= \int_0^\frac{\pi}{4} \frac{\sec^4(x)}{3\tan^4(x)+3-\sec^4(x)} \text{d}x
= \frac12 \int_0^1 \frac{t^2+1}{t^4-t^2+1} \text{d}t
$$
Which looks a little tough but reminiscent of the well-known integral:
$$ \int_0^\infty \frac{1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac12\int_0^\infty \frac{x^2+1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac{\pi}{4\cos(a)} $$
Where we choose $a=\frac\pi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?

$$3(\sin^4x+\cos^4x)-1=3((\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x)-1=2-\frac32\sin^22x=\frac{5+3\cos4x}4$$
and your integral is also
$$\int_0^{n\pi}\frac{dx}{5+3\cos x}=n\int_0^{\pi}\frac{dx}{5+3\cos x}$$ as the cosine is an even function.
Using the Weierstrass substitution, the last integral is shown to be $\dfrac\pi4$.
$$\int_0^\pi\frac{dx}{5+3\cos4x}=\int_0^\infty\frac{2\,dt}{(1+t^2)\left(5+3\dfrac{1-t^2}{1+t^2}\right)}=\int_0^\infty\frac{dt}{4+t^2}.$$
