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Evaluate the integral using the Residue Theorem. $$\int_\pi^{3\pi} \frac {dx}{5\cos x+13}$$


Residue Theorem makes my head hurt. I have a lot of trouble with Laurent series in the first place. Any help would be greatly appreciated!

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    $\begingroup$ Hint: Take advantage that $\cos$ is periodic every $2\pi$ and that $\cos(-x)=\cos(+x)$. $\endgroup$ – Simply Beautiful Art May 1 '17 at 0:18
  • $\begingroup$ to generalize SimplyBeautifulArt's comment: As a general rule of thumb, when investigating integrals (or really any mathematical object) start by looking for symmetries. Use symmetry to your advantage, and work alongside symmetry... you will often be rewarded. $\endgroup$ – Brevan Ellefsen May 1 '17 at 0:25
  • $\begingroup$ Also, to make the symmetry in this case a bit more transparent, note that the integral is the same over the intervals $[\pi,3\pi]$, $[-\pi, \pi]$ and also $[0,2\pi]$. Do you see any of these that have nice parameterizations that relate heavily to the Cosine (and Sine) functions? $\endgroup$ – Brevan Ellefsen May 1 '17 at 0:29
  • $\begingroup$ Finally, worth noting that those constants are somewhat irrelevant. It might give you some clarity to look at the antiderivative: $$\int \frac{1}{a\cos(x)+b}\mathrm{d}x = -\frac{2 \tanh ^{-1}\left(\frac{(b-a) \tan \left(\frac{x}{2}\right)}{\sqrt{a^2-b^2}}\right)}{\sqrt{a^2-b^2}}$$ $\endgroup$ – Brevan Ellefsen May 1 '17 at 0:34
  • $\begingroup$ @BrevanEllefsen Lol, the tags imply this is to be done with complex analysis, not anti-derivatives :P $\endgroup$ – Simply Beautiful Art May 1 '17 at 0:39
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The approach to evaluating this integral using contour integration is classical. It begins with the substitution $z=e^{i x}$, which implies $dx=\frac{1}{iz}\,dz$. The domain of integration transforms from $x\in [\pi,3\pi]$ to an integral on the unit circle $|z|=1$.

Proceeding as discussed, we can write

$$\begin{align} \int_\pi^{3\pi}\frac{1}{5\cos(x)+13}\,dx&=\oint_{|z|=1}\left(\frac{1}{5\left(\frac{z+z^{-1}}{2}\right)+13}\right)\,\frac{1}{iz}\,dz\\\\ &=\frac2i \oint_{|z|=1}\frac{1}{5z^2+26z+5}\,dz\\\\ &=\frac2i \oint_{|z|=1}\frac{1}{(5z+1)(z+5)}\,dz\\\\ &=2\pi i \left(\frac2i\right)\text{Res}\left(\frac{1}{(5z+1)(z+5)},z=-1/5\right)\\\\ &=\frac{\pi}{6} \end{align}$$

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  • $\begingroup$ Nice explanation and derivation. Can be used as prototype for many similar integrals. (+1) $\endgroup$ – Markus Scheuer May 1 '17 at 16:30
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    $\begingroup$ @MarkusScheuer Thank you Markus! Much appreciated. -Mark $\endgroup$ – Mark Viola May 1 '17 at 16:37
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By exploiting symmetry and the residue theorem such integral is pretty simple to tackle.
Symmetry first: $$ \int_{\pi}^{3\pi}\frac{dx}{5\cos x+13}=\int_{-\pi}^{\pi}\frac{dx}{13+5\cos(x)} = 2\int_{0}^{\pi}\frac{dx}{13+5\cos x}\\ = 2\int_{0}^{\pi/2}\frac{26}{13^2-5^2\cos^2 x}\stackrel{x\mapsto \arctan t}{=} 54\int_{0}^{+\infty}\frac{dt}{13^2(1+t^2)-5^2} $$ and the problem boils down to computing:

$$ 26\int_{-\infty}^{+\infty}\frac{dt}{12^2+13^2 t^2} = 26\cdot(2\pi i)\cdot\text{Res}\left(\frac{1}{12^2 + 13^2 t^2},t=\frac{12}{13}i\right)$$ or: $$ 26\cdot(2\pi i)\cdot\left(-\frac{i}{312}\right) = \frac{54 \pi}{312} = \color{red}{\frac{\pi}{6}}.$$

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