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?

For $x \in (0,\pi/4)$,
the integral $$\frac12 \int_0^1 \frac{t^2+1}{t^4-t^2+1} \text{d}t$$ can also be written as $$\frac12 \int_0^1 \frac{1+1/t^2}{(t-1/t)^2+1} \text{d}t$$
Substituting $u=t-\frac1t$, the integral becomes $$\frac12 \int_{-\infty}^0 \frac{1}{u^2+1} \text{d}u$$ which is equal to $$\frac12[\arctan(u)]_{-\infty}^0=\pi/4$$
Similarly for $x \in (\pi/4,\pi/2)$, the integral becomes$$\frac12 \int_1^\infty \frac{1+1/t^2}{(t-1/t)^2+1} \text{d}t$$ which is equal to$$\frac12 \int_0^{\infty} \frac{1}{v^2+1} \text{d}v$$ or$$\frac12[\arctan(u)]_0^{\infty}=\pi/4$$ 
And since the integrand has a period of $\pi/2$ because $$\sin^4x+\cos^4x=1-\frac12\sin^22x$$
one can easily contemplate the proof by adding the areas of the curve on and on.
