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I am trying to use residues to find $\int_0^\pi \frac{d\theta}{5+3\cos\theta}$.

My thoughts:

Letting $z=e^{i\theta}$ we get $dz=ie^{i\theta}$. Then, $\int_0^\pi \frac{d\theta}{5+3\cos\theta}=\frac{1}{5}\int_{|z|=1}\frac{dz}{iz(1+\frac{3}{5}(\frac{z+z^{-1}}{2}))}=-2i\int_{|z|=1}\frac{dz}{3z^2+10z+3}$.
Now, using the quadratic formula, we get that the integral becomes $-2i\int_{|z|=1}\frac{dz}{(z+3)(3z+1)}$. So, now we will compute the residue at $z=-\frac{1}{3}$ only since $z=3$ is outside of our circle. The residue is equal to $\frac{3}{8}$, and so the integral is equal to $(2\pi i)(-2i)(\frac{3}{8})=\frac{3\pi}{2}$. But, this integral, I believe should actually be equal to $\frac{\pi}{4}$ based on Wolfram.

I am wondering if I did something wrong, or (hopefully not) is it just some silly algebra mistake somewhere. Any thoughts, suggestions, etc. are always appreciated! Thank you.

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    $\begingroup$ there are two algebra mistakes - the residue has an extra $1/3$ as $\frac{z+1/3}{3z+1} \to 1/3$ and the original integral is $[0,\pi]$ so you need another $1/2$ to make it full circle (which is indeed legitimate by symmetry) $\endgroup$
    – Conrad
    Jul 12 '20 at 23:16
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    $\begingroup$ @Conrad I see that I made an algebra mistake, but shouldn't the residue be $\frac{9}{8}$ as $\lim_{z\rightarrow -\frac{1}{3}}\frac{z+\frac{1}{3}}{(3z+1)(z+3)}=\lim_{z\rightarrow -\frac{1}{3}}\frac{1}{z+3}$? $\endgroup$
    – User7238
    Jul 13 '20 at 0:06
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    $\begingroup$ $\frac{z+1/3}{3z+1} =1/3$ $\endgroup$
    – Conrad
    Jul 13 '20 at 0:10
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    $\begingroup$ Not sure what you are talking about $1/(-1/3+3)=3/8$ not $9/8$ so multiplying with the $1/3$ from the $3z+1$ vs $z+1/3$ gives you $1/8$ as the residue at $-1/3$ is just $\lim_{z\rightarrow -\frac{1}{3}}\frac{z+\frac{1}{3}}{(3z+1)(z+3)}=1/8$ $\endgroup$
    – Conrad
    Jul 13 '20 at 0:42
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    $\begingroup$ no problem - you can always double-check with the simple fraction decomposition if in doubt and here it is $1/8(3/(3z+1)-1/(z+3)$ so the limit is $1/8$ as the numerator $3$ gives precisely the denominator when multiplied by $z+1/3$ $\endgroup$
    – Conrad
    Jul 13 '20 at 0:52
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$$I = \int_0^\pi \frac{d\theta}{5+3\cos\theta}$$

I want the range to be $0$ to $2\pi$ so I will apply the substitution $\tau = \theta / 2$.

$$I = \int_0^{2 \pi} \frac{d\tau}{10+6\cos\tau}$$

Now I can apply my Residue Theorem lemma (from Freitag):

enter image description here

In our case $$f(z) = \frac{1}{i z}\frac{1}{10 + \tfrac{1}{2}6(z + \frac{1}{z})} = \frac{1}{i}\frac{1}{3z^2 + 10z + 3} = \frac{-i}{(z + 3)(3z + 1)} = \frac{-i}{3 (z + 3)(z + \tfrac{1}{3})}$$

There are two simple poles at $z_1=-3$ and $z_2=-\tfrac{1}{3}$ let's calculate the residues:

  • $\operatorname{Res}(f;z_1) = \frac{-i}{3 \cdot (z_1 + \tfrac{1}{3})} = i/8$
  • $\operatorname{Res}(f;z_2) = \frac{-i}{3 \cdot (z_2 + 3)} = -i/8$

we will only use the $z_2$ residue as $z_1$ lies outside $\mathbb E$.

So $$I = 2 \pi i \cdot - i / 8 = \pi / 4$$

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