Find $\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4x} dx$.

Question

I was wondering, how to calculate the following definite integral that appeared in an integration bee (preferably the quickest way possible).

$$\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4x} dx$$

Answer 1

Note that $$\sin^4 x + \cos^4 x = \frac{3+\cos(4x)}{4}$$ (see for example How to simplify $\sin^4 x+\cos^4 x$ using trigonometrical identities?).

Hence, by letting $z=e^{it}$, $dz=zidt$ and recalling that $\cos(t)=\frac{z+1/z}{2}$, we get \begin{align*} \int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4x} dx&= \int_{0}^{2\pi} \frac{d(4x)}{3+\cos(4x)} \\ &=\int_{0}^{8\pi}\frac{dt}{3+\cos (t)}=4\int_{0}^{2\pi}\frac{dt}{3+\cos (t)}\\ &=4 \int_{|z|=1} \frac{dz/(zi)}{3+\frac{z+1/z}{2}}\\ &=\frac{8}{i} \int_{|z|=1} \frac{dz}{z^2 + 6z + 1}\\ &=16\pi\,\mbox{Res}\left(\frac{1}{z^2 + 6z + 1},2\sqrt{2}-3\right)\\ &=16\pi\,\left.\frac{1}{(z^2 + 6z + 1)'}\right|_{z=2\sqrt{2}-3}\\ &=16\pi\,\frac{1}{2(2\sqrt{2}-3) + 6}=2\sqrt{2}\pi. \end{align*}

Answer 2

The substitution $\tan{x}=t$ helps.

$dt=\frac{1}{\cos^2x}dx=(1+t^2)dx$.

Thus, $dx=\frac{1}{1+t^2}dt$. $$\sin^4x+\cos^4x=2\cos^4x-2\cos^2x+1=\frac{2}{(1+t^2)^2}-\frac{1}{1+t^2}+1.$$ Thus, we need to calculate $$4\int\limits_0^{+\infty}\frac{\frac{1}{1+t^2}}{\frac{2}{(1+t^2)^2}-\frac{1}{1+t^2}+1}dt$$ or $$4\int\limits_0^{+\infty}\frac{1+t^2}{t^4+1}dt$$ or $$4\int\limits_0^{+\infty}\frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2}dt$$ or $$4\int\limits_{-\infty}^{+\infty}\frac{1}{x^2+2}dx,$$ which is $4\cdot\frac{\pi}{\sqrt2}=2\sqrt2\pi$.

Answer 3

Hint:

$$\sin^4x + \cos^4x = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x$$

And again

$$2\sin^2x\cos^2x = 2(\sin x \cos x)^2 = 2\left(\frac{\sin2x}{2}\right)^2 = \frac{1}{2}\sin^22x$$

Hence in the end

$$\sin^4x + \cos^4x = 1 - \frac{1}{2}\sin^22x$$

You can treat this.

Answer 4

First, note the integrand has period $\pi$ and has a symmetry w.r.t. the lines $x=\dfrac\pi2$ and $x=\dfrac\pi4$. As a consequence, $$\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4x}\,\mathrm d\mkern 2mu x= 8\int_{0}^{\tfrac\pi4} \frac{1}{\sin^4x + \cos^4x}\,\mathrm d\mkern 2mux $$

Bioche's rules suggest to set $\;t=\tan x$, so that $\;\mathrm d\mkern 2mu x=\dfrac{\mathrm d\mkern 2mu t}{1+t^2}$.

But actually, we even may set $\;u=\tan 2x$, $\;\mathrm d\mkern 2mu u=2(1+\tan^2x)\,\mathrm d\mkern 2mu x$. Indeed \begin{align} \sin^4x+\cos^4x&=(\sin^2x +\cos^2x)^2-2\sin^2x\cos^2x=1-\frac12\sin^22x\\ &=1-\frac12\frac{\tan^2 2x}{1+\tan^2 2x}=\frac{2+\tan^2 2x}{2(1+\tan^2 2x)}, \end{align} so that the integral is $$ \int_0^\infty\frac{2(1+u^2)}{2+u^2}\cdot\frac{\mathrm d\mkern 2mu u}{2(1+u^2)}= \int_0^\infty\frac{\mathrm d\mkern 2mu u}{2+u^2}=\frac1{\sqrt 2}\,\arctan\biggl(\frac u{\sqrt 2}\biggr)\biggm\vert_0^\infty=\frac\pi{2\sqrt2}, $$ and finally $$\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4x}\,\mathrm d\mkern 2mu x=2\sqrt2\,\pi.$$

Answer 5

This is related to the other answers here, but I think the logic of it is more straightforward. It proceeds in three steps.

1. Reduce the integral using the half-angle formulae and periodicity considerations. This gives $$ \int_0^{2\pi}\frac{dx}{\sin( x)^4 + \cos( x)^4} = 2\int_0^{2\pi}\frac{dx}{1+\cos(2x)^2} = 8\int_0^{\pi/2}\frac{dx}{1+\cos(2x)^2}. $$

2. Convert the integrand to a rational function using an inverse trig substitution. In this case, $2x = \mathrm{cot}^{-1}(t)$ is the preferred option, as its derivative has no square roots and simple trigonometry gives $\cos[\cot^{-1}(t)] = t^2/(1+t^2)$. With this substitution, $dx = -dt/[2(1+t^2)]$ and $$ 8\int_0^{\pi/2}\frac{dx}{1+\cos(2x)^2} = 8\int_{\infty}^{-\infty}\frac{1}{1+t^2/(1+t^2)}\left[-\frac{dt}{2(1+t^2)}\right] = 4\int_{-\infty}^\infty\frac{dt}{1+2t^2}. $$

3. Use rational function methods to evaluate the integral. I'm going to gloss over this one, though, as the integrand is clearly the derivative of an arctangent function. So we have $$ 4\int_{-\infty}^\infty\frac{dt}{1+2t^2} = \left.\frac{4}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}t\right)\right|_{-\infty}^\infty = 2\sqrt{2}\pi. $$

Thus, $$ \int_0^{2\pi}\frac{dx}{\sin( x)^4 + \cos( x)^4} = 2\sqrt{2}\pi $$