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Would someone help me to integrate

$$\int_0^{2\pi}\frac{dx}{(2+\cos(x))}$$

by means of residue theorem?

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closed as off-topic by Watson, Davide Giraudo, Cyclohexanol., Shailesh, user223391 Sep 20 '16 at 19:10

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Use $z=e^{ix}$, then $$ \begin{align} \int_0^{2\pi}\frac1{2+\cos(x)}\,\mathrm{d}x &=\oint_{|z|=1}\frac1{2+\frac{z+\frac1z}2}\frac{\mathrm{d}z}{iz}\\ &=\frac2i\oint_{|z|=1}\frac{\mathrm{d}z}{z^2+4z+1}\\ &=\frac1{i\sqrt3}\oint_{|z|=1}\left(\frac1{z+2-\sqrt3}-\frac1{z+2+\sqrt3}\right)\mathrm{d}z\\ &=\frac1{i\sqrt3}(2\pi i-0)\\[3pt] &=\frac{2\pi}{\sqrt3} \end{align} $$

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Hint. Use the substitution $z=e^{ix}$. Then $2\cos(x)=z+\frac{1}{z}$, $dz=iz\cdot dx$ and $$I:=\int_0^{2\pi}\frac{dx}{(2+\cos(x))}=\int_{|z|=1}\frac{1}{2+\frac{z+\frac{1}{z}}{2}}\cdot \frac{dz}{iz}.$$ Then apply Residue Theorem: $$I=\frac2i\int_{|z|=1}\frac{dz}{z^2+4z+1}=4\pi \cdot\mbox{Res}\left(\frac{1}{z^2+4z+1},\sqrt{3}-2\right) =\frac{4\pi}{2(\sqrt{3}-2)+4}=\frac{2\pi}{\sqrt{3}}. $$

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  • $\begingroup$ @Danijel Keno You are welcome again. Don't forget to accept and/or vote one of the answers! math.stackexchange.com/tour $\endgroup$ – Robert Z Sep 18 '16 at 11:36
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Passing into the complex plane with the substitution

$$\cos x \to \cos z = \frac{\left(z + z^{-1}\right)}{2} ~~~~~~~ \text{d}x = \frac{\text{d}z}{iz}$$

an you get:

$$\int_{|z| = 1} -i \frac{\text{d}z}{z\left(2 + \frac{z + z^{-1}}{2}\right)} = \int_{|z| = 1} -2i \frac{\text{d}z}{z\left(4 + \frac{z^2 + 1}{z}\right)}$$

Arrange the terms:

$$\int_{|z| = 1} -2i \frac{\text{d}z}{4z + z^2 + 1}$$

Now calculate the poles, solving $z^2 + 4z +1 = 0$

The first pole (solution) is $z = -2+\sqrt{3}$ inside the unit circle and it's ok.

The second one is $z = -2-\sqrt{3}$ outside the unit circle so it's not ok.

Result:

$$2\pi i\ \text{Res}_{z = -2+\sqrt{3}}\ (z - 2-\sqrt{3})\frac{-2i}{(z-2-\sqrt{3})(z-2+\sqrt{3})} = \frac{2\pi}{\sqrt{3}}$$

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  • $\begingroup$ Thank you very much for the answer and editting the question :) $\endgroup$ – Danijel Keno Sep 18 '16 at 11:43
  • $\begingroup$ @DanijelKeno Yuo're welcome! In that way your answer is clearer and cleaner :D $\endgroup$ – Von Neumann Sep 18 '16 at 11:51
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A real-analytic alternative:

$$ I = \int_{0}^{\pi}\frac{d\theta}{2+\cos\theta}+\int_{0}^{\pi}\frac{d\theta}{2-\cos\theta} = 4\int_{0}^{\pi}\frac{d\theta}{4-\cos^2\theta}=8\int_{0}^{\pi/2}\frac{d\theta}{4-\cos^2\theta}$$ and by substituting $\theta=\arctan t$ $$ I = 8 \int_{0}^{+\infty}\frac{dt}{4t^2+3}=\color{red}{\frac{2\pi}{\sqrt{3}}}. $$

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