How can one evaluate $\displaystyle\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$?

My attempt:

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

To find the singularity, I solve: $ (2+\cos\theta)^2 = 0 $ and therefore, $\cos\theta = -2$.

Substituting: $\cos z = \frac{e^{iz} + e^{-iz}}{2} = \frac{z + \frac{1}{z}}{2}$, I find that $z = -2 + \sqrt{3} $ is the singular point that lies in the unit circle $|z| = 1$.

From this point, I have little idea how to go about solving this problem. I know I have to find the residue and then just sum them but to get the expression that would cancel out the pole is where I am currently stuck.

  • 1
    $\begingroup$ If you want to do it by residues, you will have to write it as a path integral with $dz$. $\endgroup$
    – GEdgar
    Commented Feb 20, 2013 at 2:29
  • $\begingroup$ Even if $|z| \leq 1$; it doesn't mean you have $\theta \in [-\pi, \pi]$ $\endgroup$
    – hjpotter92
    Commented Feb 20, 2013 at 2:40

3 Answers 3


I will start from the point where you left off. The integral may be written as, upon substitution of $z=e^{i \theta}$:

$$\begin{align}\frac{1}{2} \int_0^{2 \pi} \frac{d\theta}{(2+\cos{\theta})^2} &= -i 2 \oint_{|z|=1} dz \frac{z}{(z^2+4 z+1)^2}\\ &= -i 2 \oint_{|z|=1} dz \frac{z}{(z+2-\sqrt{3})^2 (z+2+\sqrt{3})^2}\end{align}$$

after a bit of algebra. This integral is $i 2 \pi$ times the sum of the residues of the poles within the integration contour. The only pole inside this contour, as you point out, is the pole at $z=-2+\sqrt{3}$. The other pole at $z=-2-\sqrt{3}$ is outside this contour and is not counted.

To compute the residue at this pole, note that we have double roots. For such roots, we have to take a derivative:

$$\mathrm{Res}_{z=-2+\sqrt{3}} \frac{-i 2 z}{(z+2-\sqrt{3})^2 (z+2+\sqrt{3})^2} = \lim_{z \rightarrow -2+\sqrt{3}} \left [\frac{d}{dz} \frac{-i 2 z}{(z+2+\sqrt{3})^2} \right ]$$

I will leave the algebra to the reader; the result is $-i \sqrt{3}/9$. The integral is then

$$\int_0^{\pi} \frac{d\theta}{(2+\cos{\theta})^2} = i 2 \pi \frac{-i \sqrt{3}}{9} = \frac{2 \sqrt{3}}{9} \pi$$


Trigonometric substitution:

$$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+1}dx\;\;,\;\;\cos\theta=\frac{1-x^2}{1+x^2}\Longrightarrow$$


$$2\int\limits_0^\infty\left(\frac{1}{3+x^2}-\frac{2}{(3+x^2)^2}\right)=\left.\frac{2}{\sqrt 3}\arctan\frac{x}{\sqrt 3}\right|_0^\infty-\left.4\left(\frac{x}{6(3+x^2)}+\frac{1}{6\sqrt 3}\arctan\frac{x}{\sqrt 3}\right)\right|_0^\infty=$$

$$=\frac{4}{3\sqrt 3}\frac{\pi}{2}=\frac{2\pi}{3\sqrt 3}$$


$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\pi}{\dd\theta \over \bracks{2 + \cos\pars{\theta}}^{\,2}}} = \left.-\,\totald{}{\mu}\int_{0}^{\pi}{\dd\theta \over \mu + \cos\pars{\theta}} \right\vert_{\ \mu\ =\ 2} \\[5mm] = &\ -\,\totald{}{\mu}\braces{\int_{0}^{\pi/2}{\dd\theta \over \mu + \cos\pars{\theta}} + \int_{-\pi/2}^{0}{\dd\theta \over \mu - \cos\pars{\theta}}}_{\mu\ =\ 2} \\[5mm] = &\ \left.-\,\totald{}{\mu}\int_{0}^{\pi/2}{2\mu \over \mu^{2} - \cos^{2}\pars{\theta}}\,\dd\theta\,\right\vert_{\ \mu\ =\ 2} \\[5mm] = &\ \left.-2\,\totald{}{\mu}\int_{0}^{\pi/2}{\mu \over \mu^{2}\sec^{2}\pars{\theta} - 1} \,\sec^{2}\pars{\theta}\,\dd\theta\,\right\vert_{\ \mu\ =\ 2} \\[5mm] = &\ \left.-2\,\totald{}{\mu}\int_{0}^{\infty}{\mu \over \mu^{2}t^{2} + \mu^{2} - 1} \,\dd t\,\right\vert_{\ \mu\ =\ 2} \\[5mm] = & \left.-\pi\,\totald{}{\mu}{1 \over \root{\mu^{2} - 1}}\right\vert_{\ \mu\ =\ 2} = \bbx{\ds{{2\root{3} \over 9}\,\pi}} \end{align}

  • $\begingroup$ this answer is really good, the only reason why it gets only two upvotes is being late in the party $\endgroup$
    – Shing
    Commented May 15, 2019 at 13:11
  • 1
    $\begingroup$ @Shing That's true. Sometimes old questions reappear in the timeline due to a MSE-Algorithm ( I guess so ). Thanks. $\endgroup$ Commented May 15, 2019 at 20:20

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