# Using a circular contour integral

I was having some problems preparing for an exam, and a friend of mine told me about this site :)

I have to prove this: $$\int_0^{2\pi} \frac{d\theta}{a + \cos\theta} = \frac{2\pi}{\sqrt{a^2 - 1}}$$

Using

$$z = e^{i\theta}\\ a>1$$

and integrating over the unit circle $|z| = 1$.

I know there are proofs of this relationship, but I can't manage to do it using the unit circle contour.

Afterwards I also have to proof a similar relation, with the integrand squared: $$\int_0^{2\pi} \frac{d\theta}{( a + cos\theta)^2} = \frac{2a\pi}{(a^2 - 1)^{3/2}}$$

I've tried to put up the equations, but as far as I can tell there are no poles ($z = -a$ lies outside of the unit circle ). Then I can rewrite

$$\frac{1}{a + z} = \frac{1}{a + \cos\theta + i\sin\theta}$$

But then I'm stuck :(

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concerning your second integral the result is immediate if you use derivation of $a$ under the integral sign, –  Raymond Manzoni Aug 10 '12 at 11:33

Write \begin{align} \frac 1{a+\cos\theta}&=\frac 2{2a+e^{i\theta}+e^{-i\theta}}\\ &=\frac{2e^{i\theta}}{e^{2i\theta}+2ae^{i\theta}+1}\\ &=\frac 2i\frac{ie^{i\theta}}{e^{2i\theta}+2ae^{i\theta}+1}, \end{align} and integrating \begin{align} \int_0^{2\pi}\frac 1{a+\cos\theta}&=\frac 2i\int_{C(0,1)}\frac 1{z^2+2az+1}dz\\ &=\frac 2i\frac 1{2\sqrt{a^2-1}}\int_{C(0,1)}\left(\frac 1{z-(a-\sqrt{a^2-1})}-\frac 1{z-(a+\sqrt{a^2-1})}\right)dz\\ &=\frac 1{2\pi i}\frac{2\pi}{\sqrt{a^2-1}}\int_{C(0,1)}\left(\frac 1{z-(a-\sqrt{a^2-1})}-\frac 1{z-(a+\sqrt{a^2-1})}\right)dz. \end{align} Now we use Cauchy's integral formula, noting that $a+\sqrt{a^2-1}$ is outside the unit circle.

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$$I= \int _0^{2\pi} \frac{d\theta}{a + \cos\theta}$$

$$=2\int _0^{\pi} \frac{d\theta}{a + \cos\theta}$$

as $\int _0^{2\pi} \frac{d\theta}{a + \cos\theta}=\int _0^{\pi} \frac{d\theta}{a + \cos\theta}+\int _\pi^{2\pi} \frac{d\theta}{a + \cos\theta}$

Now , $\int _\pi^{2\pi} \frac{d\theta}{a + \cos\theta}=-\int _\pi ^{0} \frac{dy}{a + \cos y}$ where $y=2\pi-\theta$

So, $\int _\pi^{2\pi} \frac{d\theta}{a + \cos\theta}=\int _0^\pi \frac{dy}{a + \cos y}$

$I= 2\int _0^{\pi} \frac{d\theta}{a + \frac{(1-tan^2\frac{\theta}{2})}{(1+tan^2\frac{\theta}{2})}}$

$= 2\int _0^{\pi} \frac{sec^2\frac{\theta}{2}d\theta}{(a+1) + (a-1)tan^2\frac{\theta}{2}}$

Now putting $tan\frac{\theta}{2}=z$, $sec^2\frac{\theta}{2}d\theta=2dz$

$=\int _0^∞ \frac{4dz}{(a+1) + (a-1)z^2}$

$=\frac{4}{a-1}\int _0^∞\frac{dz}{\frac{a+1}{a-1} + z^2}$

$=\frac{4}{a-1}\sqrt{\frac{a-1}{a+1}}{}tan^{-1}(\frac{z(a-1)}{a+1})| _0^∞$

$=\frac{4}{\sqrt{a^2-1}}\frac{\pi}{2}$

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