Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $

Question

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?

Answer 1

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.

Answer 2

$$\int_0^{\pi/4}\frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{\pi}{4}$$ can be shown by using partial fraction in OP's substituted integral. One can similarly show $$\int_{\pi/4}^{\pi/2}\frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{\pi}{4}$$ Observe that the integrand has period $\frac\pi2$, the problem can be easily done by using induction.

Answer 3

Note that the integrand function $f(x)=\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}$ is even with period $\pi/2$, so $$\int_0^\frac{n\pi}{4} f(x) dx= \frac{1}{2}\int_{-\frac{n\pi}{4}}^\frac{n\pi}{4} f(x) dx= \frac{1}{2}\int_{0}^\frac{n\pi}{2} f(x) dx =\frac{n}{2}\int_{0}^\frac{\pi}{2} f(x) dx.$$ Now, following your approach, we have that $$\frac{t^2+1}{t^4-t^2+1} = \frac{2}{1+(2t+\sqrt{3})^2} +\frac{2}{1+(2t-\sqrt{3})^2}.$$ Therefore $$\int\frac{t^2+1}{t^4-t^2+1}\,dt=\arctan(2t+\sqrt{3})+\arctan(2t-\sqrt{3})+c$$ Hence, for $t=\tan(x)$, $$\int_0^\frac{\pi}{2} f(x) dx = \frac12 \int_0^{\infty} \frac{t^2+1}{t^4-t^2+1} dt=\frac{1}{2}\left[\arctan(2t+\sqrt{3})+\arctan(2t-\sqrt{3})\right]_0^{+\infty}=\frac{\pi}{2}.$$

Answer 4

$$3(\sin^4x+\cos^4x)-1=3((\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x)-1=2-\frac32\sin^22x=\frac{5+3\cos4x}4$$

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 $\dfrac\pi4$.

$$\int_0^\pi\frac{dx}{5+3\cos4x}=\int_0^\infty\frac{2\,dt}{(1+t^2)\left(5+3\dfrac{1-t^2}{1+t^2}\right)}=\int_0^\infty\frac{dt}{4+t^2}.$$