# Evaluating a Complex Trigonometric Integral

This is from a practice exam that we are working on, problem number 2. We were thinking first to use Cauchy-Goursat, but then the problem only says that the curve doesn't lie on the singularities of $\sin^2(z)$. They might still be interior to the curve. Is there another way to approach the problem using theorems we might know?

Edit: The question is : Prove that $$\int_C\frac{\cos(z)\mathrm dz}{\sin^2(z)}=0$$ where C is any simple closed contour not passing through a zero of $\sin(z)$.
Edit: I don't know about residues.

• You should post the question also instead of linking to the pdf file. – Adrián Barquero Apr 1 '11 at 3:19
• okay. I will post the question as well. – I Love Cake Apr 1 '11 at 4:26
• Should that be $\sin^2(z)$ instead of $\sin^2(x)$? $x$ hasn't appeared before. – Ross Millikan Apr 1 '11 at 4:33
• Sorry about that. Let me fix that. – I Love Cake Apr 1 '11 at 4:33

## 2 Answers

You might notice that $\frac{\cos(z)}{\sin^2(z)}$ has an antiderivative (which any calculus student should be able to find) on the complement of its singularities. You have probably seen a theorem that talks about contour integrals of a function that has an antiderivative.

Do you know about residues? If you do, notice that the integrand is a meromorphic function all of whose poles are double with zero residue. Using that, the result is immediate.

• I don't know about residues actually. – I Love Cake Apr 1 '11 at 5:28
• @ILoveCake: well... it would help if you expanded the question with a description of what you do know :) – Mariano Suárez-Álvarez Apr 1 '11 at 5:30