1. ## Integral

Hi, I'm really stuck on this question, I don't really know where to start.

Calculate the integral $\int_0^\pi \displaystyle\frac{1}{(\alpha+cos\theta)^2}d\theta$ where $\alpha>1$

In my notes I have similar-ish examples which used the substitution $z=e^{i\theta}$ but in those, the integral went from 0 to $2\pi$, so gave a closed contour & then we could use residues. I thought maybe I could substitute $z=e^{2i\theta}$ instead but that didn't seem to get me anywhere. So now I'm clueless, any hints would be much appreciated.

2. Notice that since cosine is an even function, the integrand is an even function and thus $\displaystyle I = \int_0^\pi \displaystyle\frac{1}{(\alpha+cos\theta)^2}d\theta = \frac{1}{2} \int_{- \pi}^\pi \displaystyle\frac{1}{(\alpha+cos\theta)^2}d\theta$

3. Originally Posted by roshx
Hi, I'm really stuck on this question, I don't really know where to start.

Calculate the integral $\int_0^\pi \displaystyle\frac{1}{(\alpha+cos\theta)^2}d\theta$ where $\alpha>1$

In my notes I have similar-ish examples which used the substitution $z=e^{i\theta}$ but in those, the integral went from 0 to $2\pi$, so gave a closed contour & then we could use residues. I thought maybe I could substitute $z=e^{2i\theta}$ instead but that didn't seem to get me anywhere. So now I'm clueless, any hints would be much appreciated.
The first thing to notice is that the integral is symmetric over the vertical line $x=\pi$

This gives $
\frac{1}{(\alpha+cos\theta)^2}d\theta$

Too slow
$\displaystyle \int_{0}^{\pi}\frac{1}{(\alpha+cos\theta)^2}d\thet a=\frac{1}{2}\int_{0}^{2\pi}\frac{1}{(\alpha+cos\t heta)^2}d\theta$

Now use the sub you suggested before!

4. Hey, thanks for the help everyone. I've been working on this problem and I've got a bit stuck again. I did the substitution and took out constants, now have the integrand looking like this:

$\displaystyle\frac{1}{z(\alpha + \frac{1}{2}(z+\frac{1}{z}))^2}$

I think z=0 is the only singularity I need to worry about since any others would be outside the contour (the unit circle) but please correct me if I'm wrong.

However I can't seem to work out the residue. I keep winding up with a pesky 1/z. Is z=0 a simple pole?