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

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.

• If you want to do it by residues, you will have to write it as a path integral with $dz$. Feb 20 '13 at 2:29
• Even if $|z| \leq 1$; it doesn't mean you have $\theta \in [-\pi, \pi]$ Feb 20 '13 at 2:40

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:

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

$$\int\limits_0^\pi\frac{d\theta}{(2+\cos\theta)^2}=\int\limits_0^\infty\frac{2\,dx}{1+x^2}\frac{1}{\left(2+\frac{1-x^2}{1+x^2}\right)^2}=2\int\limits_0^\infty\frac{x^2+1}{(x^2+3)^2}dx=$$

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

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


• this answer is really good, the only reason why it gets only two upvotes is being late in the party May 15 '19 at 13:11
• @Shing That's true. Sometimes old questions reappear in the timeline due to a MSE-Algorithm ( I guess so ). Thanks. May 15 '19 at 20:20