Deon Moran

2022-10-24

Where the method used should be using complex analysis.
${\int }_{c}\frac{d\theta }{\left(p+\mathrm{cos}\theta {\right)}^{2}}=\frac{2\pi p}{\left({p}^{2}-1\right)\sqrt{{p}^{2}-1}};c:|z|=1$

Kash Osborn

i do my self like this on $|z|=1,z={e}^{i\theta },d\theta =\frac{dz}{iz}$
using substitution $\mathrm{cos}\theta =\frac{z+{z}^{-1}}{2}$
$\frac{1}{i}\int \frac{\frac{dz}{z}}{\left(p+\frac{z+{z}^{-1}}{2}\right)\left(p+\frac{z+{z}^{-1}}{2}\right)}or\frac{1}{i}\int \frac{1}{\left(2pz+{z}^{2}+1\right)\left(p+\frac{z+{z}^{-1}}{2}\right)}$
and then multiple by $\frac{z}{z}$
$\frac{1}{i}\int \frac{z}{\left(2pz+{z}^{2}+1\right)\left(2pz+{z}^{2}+1\right)}$
the roots of $\left(2pz+{z}^{2}+1\right)$, ${z}_{1}=-p+\sqrt{{p}^{2}-1};{z}_{2}=-p-\sqrt{{p}^{2}-1}$
let ${z}_{1}=a$ and ${z}_{2}=b$
so $a-b=2\sqrt{{p}^{2}-1}$ and $a+b=-2p$
and then i use residue theorm, but the result on the right $\frac{\pi p}{2\left({p}^{2}-1\right)\sqrt{{p}^{2}-1}}$

Christopher Saunders

On $|z|=1$, we have $z={e}^{i\theta }$, so $\frac{dz}{d\theta }=i{e}^{i\theta }$, and hence $d\theta =\frac{dz}{iz}$
Also on |z|=1, we have $\mathrm{cos}\theta =\frac{1}{2}\left(z+\frac{1}{z}\right)$
Putting this together, you get

Expand the denominator,

Do you have a similar question?