# Evaluating the contour integral: $\oint_C \frac{\sin 2z}{(6z-\pi)^3}dz$

I am trying to evaluate the following integral, but don't know how to take the coefficient of $z$ out of the parenthesis to get it into the Cauchy integral form. Any help is appreciated.

$$\oint_C \frac{\sin 2z}{(6z-\pi)^3}dz$$

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## 1 Answer

Do you mean as in $$6^3\left(z-\frac{\pi}{6}\right)^3?$$

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Hmm. I tried expanding it out and got into a mess, and didn't realize it was that simple. Thanks. –  Joebevo Jul 17 '12 at 6:38
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