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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.

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    Sorry about that. Let me fix that.2011-04-01

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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.

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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.

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    @ILoveCake: well... it would help if you expanded the question with a description of what you do know :)2011-04-01