# Complex analysis - evaluate integral

#### matteoit81

##### New member
Hi all,

I need to evaluate this integral

anybody could point me to a solution?
I've tried to look around (google, books), but I found no clue to solve it

I wrote it in latex

$\displaystyle \int_0^{2\pi} \! \frac{1}{(2 + \cos \theta)^2} \mathrm{d} \theta$

Thanks for the help,
matteo

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#### matteoit81

##### New member
Thank for the fast rea\ply. I didn't know about mathematica website, very useful!!

There is just one "problem"

I read the solution of the integral and it is complete but way too complicated...
this one is a typical proposed exercise of complex analysis course and I don't think we are supposed to evaluate it in this way

the course topics are related to taylor/laurent series or residue calculus

is there any easier way to compute the defined integral?

#### chisigma

##### Well-known member
Thank for the fast rea\ply. I didn't know about mathematica website, very useful!!

There is just one "problem"

I read the solution of the integral and it is complete but way too complicated...
this one is a typical proposed exercise of complex analysis course and I don't think we are supposed to evaluate it in this way

the course topics are related to taylor/laurent series or residue calculus

is there any easier way to compute the defined integral?
... and there is one more minor problem: the integral to be computed is...

$\displaystyle \int_{0}^{2 \pi} \frac{d\ \theta}{(2+\cos \theta)^{2}}$ (1)

... where the denominator is squared and not $\displaystyle \int_{0}^{2 \pi} \frac{d\ \theta}{2+\cos \theta}$!...

The standard procedure to compute an integral like (1) using complex analysis is to set $\displaystyle z=e^{i\ \theta} \implies \cos \theta= \frac{z+z^{-1}}{2}\ ,\ d \theta= -i\ \frac{d z}{z}$, so that the integral becomes...

$\displaystyle \int_{0}^{2 \pi} \frac{d\ \theta}{(2+\cos \theta)^{2}}= -16\ i\ \int_{C} \frac {z}{z^{4}+8 z^{3} + 18 z^{2} +8 z +1}\ dz$ (2)

... where C is the unit circle [a circle centered in z=0 and with radious 1...]. The f(z) in (2) has a pair of poles with multiplicity 2 in $\displaystyle z_{1}= -2 -\sqrt{3}$ and $z_{2}= -2+\sqrt{3}$ and only the last is inside the unit circle. Applying the Cauchy integral theorem the (2) is $\displaystyle I= 2\ \pi\ i\ r_{1}$ where...

$\displaystyle r_{1}= \lim_{z \rightarrow z_{2}}\frac{d}{dz}\ \{ f(z)\ (z-z_{2})^{2} \}= -16\ i\ \lim_{z \rightarrow z_{2}} \frac{d}{dz} \frac{z}{(z+2+\sqrt{3})^{2}}$ (3)

The details of computation are tedious but not too difficult and are left to 'Matteo'...

Kind regards

$\chi$ $\sigma$

#### Sherlock

##### Member
Two words: 'magic differentiation'. Define:

\begin{aligned}I(\lambda) & :=\int_{0}^{2\pi}\frac{1}{\lambda+a\cos{x}}\;{dx} \\& =\int_{0}^{2\pi}\frac{1}{\lambda\left(\sin^{2} \frac{1}{2}x+\cos^{2} \frac{1}{2}x\right)+a\left(\cos^{2} \frac{1}{2}x-\sin^{2}\frac{1}{2}x\right)}\;{dx}\\& =\int_{0}^{2\pi}\frac{1}{( \lambda-a)\sin^{2}\frac{1}{2}x+(\lambda+a)\cos^{2} \frac{1}{2} x}\;{dx}\\& =\int_{0}^{2\pi}\frac{\sec^{2}{\frac{1}{2}x}}{( \lambda+a)+( \lambda-a)\tan^{2}\frac{1}{2}x}\;{dx}\\& = 4\int_{0}^{\infty}\frac{1}{( \lambda+a)+(\lambda-a)t^{2}}\;{dt}\\& = 4\int_{0}^{\infty}\frac{1}{(\sqrt{\lambda+a})^{2}+(\sqrt{\lambda-a})^{2}t^{2}}\;{dt}\\& =\frac{4}{{\sqrt{\lambda^{2}-a^{2}}}}\tan^{-1}\bigg(\frac{\sqrt{\lambda-a}}{\sqrt{\lambda+a}}~t\bigg)\bigg|_{0}^{\infty}\\& =\frac{2\pi}{\sqrt{\lambda^{2}-a^{2}}}.\end{aligned}

But $\displaystyle I'(\lambda) =-\int_{0}^{2\pi}\frac{1}{(\lambda+a\cos{x})^{2}}\;{dx}$, thus:

$\displaystyle \int_{0}^{2\pi}\frac{1}{(\lambda+a\cos{x})^{2}}\;{dx}=\frac{2\lambda\pi}{\sqrt{(\lambda^{2}-a^{2})^{3}}}.$