Help me to prove this integration

Question

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

Answer 1

i do my self like this on $|z| = 1, z = e^{i\theta}, d\theta=\frac{dz}{iz}$

using substitution $\cos\theta=\frac{z+z^{-1}}{2}$

$\frac{1}{i}\int\frac{\frac{dz}{z}}{(p+\frac{z+{z}^{-1}}{2})(p+\frac{z+{z}^{-1}}{2})} or \frac{1}{i}\int\frac{1}{(2pz+z^2+1)(p+\frac{z+{z}^{-1}}{2})} $

and then multiple by $\frac{z}{z}$

$\frac{1}{i}\int\frac{z}{(2pz+z^2+1)(2pz+z^2+1)}$

the roots of $(2pz+z^2+1)$, $z_{1}=-p+\sqrt{p^2-1};z_{2}=-p-\sqrt{p^2-1}$

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 $\frac{\pi p}{2(p^2-1)\sqrt{p^2-1}}$

Answer 2

Hint: Let $z=e^{i \theta}$. Then $d\theta = dz/(i z)$ and $\cos{\theta} = (1/2) (z+(1/z))$. You should get a polynomial in the denominator in $z$ - the zeroes of these are the poles of the integrand. Then use the Residue theorem to express the integral in terms of the residues of those poles within $C$.

Answer 3

Hint: $$\cos \theta = \frac {z + z^{-1}} 2$$

Plug this in and refer to classic tools in complex analysis, such as the Cauchy formula or the residue theorem.

Answer 4

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 $\cos \theta = \frac12\left(z+\frac{1}{z}\right)$.

Putting this together, you get

$$\frac{1}{i} \int_{|z| = 1} \frac{1}{z\left(p+\frac12\left(z+\frac{1}{z}\right)\right)^2}\ dz.$$

Expand the denominator,

$$\begin{align*} \frac{1}{i} \int_{|z| = 1} \frac{1}{z\left(p+\frac12\left(z+\frac{1}{z}\right)\right)^2}\ dz &= \frac{1}{i} \int_{|z| = 1} \frac{1}{z\left(p^2 + p(z+1/z) + \frac14(z+1/z)^2\right)} \\ &= \frac{1}{i} \int_{|z| = 1} \frac{1}{\left(p^2z + pz^2+p + \frac{z}{4}(z^2+2+1/z^2) \right)} \\ &= \cdots \end{align*} $$

Then, apply the residue theorem.