∫dθ/(1 + (sinθ)^2 ), [0, π]
Cauchy's Formula, perhaps Cauchy's Thm.
The Attempt at a Solution
I first substituted sinθ = (1/(2i))(z - 1/z), where z = e^(2iθ) to get the expression
∫4iz/(z^4 - 6z^2 + 1)dz with the path |z| = 1.
I found the zeros for the denominator, which are z^2 = 3 ± 2√(2), and so z = ±√(3 ± 2√(2)). Only the points ±√(3 - 2√(2)) are in |z|= 1.
I proceeded to use Cauchy's formula to get 0, however, I am not confident with my answer. There are several things that I did as a result of my unfamiliarity of domains other than [0, 2π] when it comes to application of Cauchy's formula. I am not sure I am allowed to use the substitute of z but if I don't then I do not believe we are considering a closed curve at all.
By the way, I am trying to find an answer in a specific way, I know there are other ways to solve this problem but my book says it can be done using something Cauchy related. I hope my question is well stated and thank you.