- #1

- 28

- 0

## Homework Statement

Evaluate the integral:

[itex]\int[/itex][itex]^{2\pi}_{0}[/itex][itex]\frac{d\theta}{(A+Bcos(\theta))^2}[/itex]

[itex]a^2>b^2[/itex]

[itex]a>0[/itex]

## The Attempt at a Solution

First, I convert this to contour integration along a full sphere in the complex plane.

I let:

[itex]z=e^(i\theta)[/itex]

[itex]dz=ie^(i\theta)[/itex]

[itex]d\theta=-idz/z[/itex]

[itex]cos(\theta)=(z+z^-1)/2[/itex]

Now, substituting back into the integral, I drop the integral sign for now and just work on the integrand:

=[itex]\frac{-idz}{z(a+\frac{bz+bz^-1}{2})}[/itex]

=[itex]\frac{-idz}{z(\frac{2a+bz+bz^-1}{2})}[/itex]

=[itex]\frac{-4idz}{z(2a+bz+bz^-1)^2}[/itex]

=[itex]\frac{-4idz}{b^2z^3+4abz^2+4a^2z+2b^2z+4ab+\frac{b^2}{z}}[/itex]

And from here I am stuck on how to find the poles of this function. I want to use the residue theorem to evaluate this integral, but like I said I'm stuck here.

Last edited: