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

Question

I''m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. Here it goes:

We want to evaluate $\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta$.

Here's my work:

Let $z=e^{i\theta}$ so that $d\theta=dz/iz$ and $\cos \theta = \frac12 (z+z^{-1})$. We have

$$\begin{align} I&=\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta \quad (1)\\\\ &=\oint_C \frac{1}{(2+\frac{z+z^{-1}}{2})^2}\frac{dz}{iz} \quad (2)\\\\ &=\frac1i\oint_C \frac{4}{(4+z+z^{-1})^2}\frac{dz}{z} \quad (3) \\\\ &=\frac4i\oint_C \frac{1}{(z^2+4z+1)^2}\,dz \quad (4) \\\\ &=\frac4i\oint_C \frac{1}{(z-(-2+\sqrt3)^2(z-(-2-\sqrt3)^2}\,dz \quad (5)\\\\ &=\frac4i\oint_Cf(z)\,dz \quad (6) \end{align}$$

where $C$ is the unit circle in the complex $z$-plane.

The function $f(z)$ has singularities at $(-2\pm\sqrt3)$ but only $(-2+\sqrt3)\in int(C)$. Therefore,

$$\oint_Cf(z)\,dz=2\pi i (Res(f, (-2+\sqrt3))) \quad (*)$$

Since $(-2+\sqrt3)$ is a pole of order $2$, after a quick calculation I get

$$Res(f, (-2+\sqrt3)=-\frac{\sqrt3}{3}$$

and so for $(*)$ I get

$$\oint_Cf(z)\,dz=2\pi i (-\frac{\sqrt3}{3})=-\frac{2\pi\sqrt3i}{3}.$$

Hence, by $(6)$, I find

$$\frac4i\oint_Cf(z)\,dz=\frac4i \left(\frac{-2\pi\sqrt3i}{3}\right)=-\frac{8\pi\sqrt3}{3}.$$

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My solution however is not correct. The given integral evaluates to $\frac{4\pi}{3\sqrt3}$ (verified with Wolfram). I suspect there must be a flaw in one of my steps from $(1)$ to $(6)$ but I can't find it.

Answer 1

Note that $(4)$ is $$\frac{4}{i} \oint_{C}\frac{\color{red}{z}}{\left(z^{2}+4z+1\right)^{2}}dz $$ and $$\textrm{Res}_{z=-2+\sqrt{3}}\left(\frac{z}{\left(z^{2}+4z+1\right)^{2}}\right)=\frac{1}{6\sqrt{3}}.$$

Answer 2

Edit $$I(a)=\int_{0}^{2\pi}\frac{1}{a+\cos x}dx=\int_{-\pi}^{\pi}\frac{1}{a-\cos x}dx=2\int_{0}^{\pi}\frac{1}{a-\cos x}dx$$ We know $\cos x=\large \frac{1-\tan^2\frac x 2}{{1-\tan^2\frac x 2}}$,thus $$I(a)=2\int_{0}^{\pi}\frac{{1+\tan^2\frac x 2}}{(a-1)+(a+1)\tan^2\frac{x}{2}}dx=\frac{2}{(a+1)}\int_{0}^{\pi}\frac{1+\tan^2\frac x 2}{\frac{a-1}{a+1}+\tan^2\frac{x}{2}}dx$$ set $u=\tan\frac{x}{2}$, we have $$I(a)=\frac{4}{a+1}\int_{0}^{+\infty}\frac{1}{\frac{a-1}{a+1}+u^2}du=\frac{4}{a+1}\sqrt{\frac{a+1}{a-1}}\tan^{-1}\left(\frac{a+1}{a-1}u\right)\large|_{0}^{\infty}=\frac{2\pi}{\sqrt{a^2-1}}$$

$$I(a)=\int_{0}^{2\pi}\frac{1}{a+\cos x}dx=\frac{2\pi}{\sqrt{a^2-1}}$$

$$I'(a)=-\int_{0}^{2\pi}\frac{1}{(a+\cos x)^2}dx=-\frac{2\pi a}{(a^2-1)^\frac32}$$ In the other words

$$\int_{0}^{2\pi}\frac{1}{(a+\cos x)^2}dx=\frac{2\pi a}{(a^2-1)^\frac32}$$

now set $a=2$