Contour Integration used to solved real Integrals

[Jufro]

Homework Statement

Essential Mathematical Methods for the Physics Sciences Problem 15.7

Show that if f(z) has a simple zero at z₀ then 1/ f(z) has a residue of 1/f'(z₀). Then use this information to evaluate:

∫ sinθ/(a- sinθ) dθ, where the integral goes from -∏ to ∏.

Homework Equations

The book has further hints:
The unit circle has only one pole at z= i*a-i(a²-1)^1/2 and therefore has a residue of -i/2 *(a² -1)^-1/2.

The Attempt at a Solution

The first part is:
1/f(z) has a simple pole at z₀ leaving the residue to be lim_z→z0 (z-z₀) * 1/f(z) = 1/f'(z₀) by definition of the derivative.

But using this information to do the integral is escaping me. Can anyone just help me understand how to use the hint. I don't actually need the whole step by step guide to the integral.

 

[vela]

I haven't worked it out, but I'd try using the substitution $z=e^{i\theta}$ and expressing $\sin\theta$ as $\frac{z-1/z}{2i}$.

 

[jackmell]

Homework Statement

Essential Mathematical Methods for the Physics Sciences Problem 15.7

Show that if f(z) has a simple zero at z₀ then 1/ f(z) has a residue of 1/f'(z₀). Then use this information to evaluate:

∫ sinθ/(a- sinθ) dθ, where the integral goes from -∏ to ∏.

Homework Equations

The book has further hints:
The unit circle has only one pole at z= i*a-i(a²-1)^1/2 and therefore has a residue of -i/2 *(a² -1)^-1/2.

The Attempt at a Solution

The first part is:
1/f(z) has a simple pole at z₀ leaving the residue to be lim_z→z0 (z-z₀) * 1/f(z) = 1/f'(z₀) by definition of the derivative.

But using this information to do the integral is escaping me. Can anyone just help me understand how to use the hint. I don't actually need the whole step by step guide to the integral.

What is a? That matters. For example if |a|<1 and real, the integral becomes a principal-valued integral.

 

[Jufro]

Sorry, a is real and a > 1. Rewriting sin as vela said came to the right answer, thanks.