Integration of complex functions with trig functions: $\int_0^{2 \pi} \frac{ d\theta}{5-\cos( \theta )}$

Question

$\int_0^{2 \pi} \frac{ d\theta}{5-\cos( \theta )}$

How should I integrate this? Using the exponantial identities of trig? Any hints will be great...

Thank you!

Answer 1

$$I=\int_0^{2\pi}\frac{d\theta}{5-\cos(\theta)}$$

let $e^{i\theta}=z$

$$I=\int_{|z|=1}\frac{1}{5-\frac{z+\frac{1}{z}}{2}}\frac{dz}{iz}$$

$$I=i\int_{|z|=1}\frac{2}{z^2-10z+1}dz$$

the root of $z^2-10z+1=0$

$z_1=5+2\sqrt 6$

$z_2=5-2\sqrt6$

you notice that $z_1$ outside of the unit circle

and $z_2$ inside the unit circle

so be the residue theorem

$$I=i\text{Res}_{z=z_2}\frac{2}{(z-z_1)(z-z_2)}=\frac{\pi}{\sqrt6}$$

Answer 2

$$\begin{eqnarray*}\int_{0}^{2\pi}\frac{d\theta}{5-\cos\theta}&=&2\int_{0}^{\pi}\frac{d\theta}{5-\cos\theta}=2\int_{0}^{\pi}\frac{d\theta}{5-\frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}}=4\int_{0}^{\pi/2}\frac{d\theta}{5-\frac{1-\tan^2\theta}{1+\tan^2\theta}}\\&=&4\int_{0}^{+\infty}\frac{dt}{5(1+t^2)-(1-t^2)}=\int_{0}^{+\infty}\frac{dt}{1+\frac{3}{2}t^2}=\color{red}{\frac{\pi}{\sqrt{6}}}.\end{eqnarray*}$$

Answer 3

Here's an answer using Fourier series.

First fact: if $\beta\in(-1,1)$, $$\sum_{n=0}^{+\infty}\beta^n\cos(n\theta)=\Re\left(\frac1{1-\beta\mathrm{e}^{i\theta}}\right)=\frac{1-\beta\cos\theta}{1+\beta^2-2\beta\cos\theta}=\frac12\frac{2-2\beta\cos\theta}{1+\beta^2-2\beta\cos\theta}=\frac12\frac{\bigl(1+\beta^2-2\beta\cos\theta\bigr)+1-\beta^2}{1+\beta^2-2\beta\cos\theta}=\frac12+\frac12\frac{1-\beta^2}{1+\beta^2-2\beta\cos\theta}.$$

Now we find $\beta\in(-1,1)$ such that $1+\beta^2=10\beta$ so that the denominator is a multiple of $5-\cos\theta$: $$\beta=5-2\sqrt6$$ fulfills this requirement. Observe that $1-\beta^2=2-10\beta$, we'll use it below.

With this value of $\beta$ we have:

$$\sum_{n=0}^{+\infty}\beta^n\cos(n\theta)=\frac12+\frac1{2\beta}\frac{1-5\beta}{5-\cos\theta}.$$

Now, integrating on a period, we clearly have: $$1=\frac1{2\pi}\int_0^{2\pi}\left(\frac12+\frac1{2\beta}\frac{1-5\beta}{5-\cos\theta}\right)\,\mathrm{d}\theta,$$ i.e., $$1=\frac12+\frac{1-5\beta}{4\beta\pi}\int_0^{2\pi}\frac1{5-\cos\theta}\,\mathrm{d}\theta,$$ hence $$\int_0^{2\pi}\frac1{5-\cos\theta}\,\mathrm{d}\theta=\frac{2\beta\pi}{1-5\beta}=\frac{\pi\sqrt6}6.$$

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As a by-product, we also obtain a more general formula: for all $n\in\mathbb{N}^*$, $$\beta^n=\frac1\pi\int_0^{2\pi}\left(\frac12+\frac1{2\beta}\frac{1-5\beta}{5-\cos\theta}\right)\cos(n\theta)\,\mathrm{d}\theta,$$ hence $$\beta^n=\frac1{2\beta\pi}\int_0^{2\pi}\frac{1-5\beta}{5-\cos\theta}\cos(n\theta)\,\mathrm{d}\theta,$$ hence $$\frac{2\beta^{n+1}\pi}{1-5\beta}=\int_0^{2\pi}\frac{\cos(n\theta)}{5-\cos\theta}\,\mathrm{d}\theta,$$ (and this equality also happens to be correct for $n=0$).