Slightly Harder Cauchy Integral

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    Cauchy Integral
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SUMMARY

The integral I_1 = ∫_0^{2π} (dθ)/(5-3sinθ)^2 can be evaluated using complex analysis techniques. By substituting z = e^{iθ}, the integral transforms into I_2 = ∫_0^{2π} (1/(5-3z)^2)(dz/(iz)), which reveals a simple pole at z=0 and a second-order pole at z=5/3. The evaluation of the integral requires calculating the residues at these poles, with the contour of integration typically being the unit circle due to the nature of the substitution. The discussion emphasizes the importance of understanding residue theory in complex analysis for solving such integrals.

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Homework Statement



Evaluate the integral I_1 = \int_0^{2\pi} \frac{d\theta}{(5-3sin\theta)^2}

Homework Equations





The Attempt at a Solution



I start off by switching the sine term for a complex exponential e^{i\theta}=cos\theta +isin\theta
I will consider only the Imaginary component of the solution.

now make the substitution: z=e^{i\theta}

so we have:

I_1 = Im\left(I_2\right)

I_2 = \int_0^{2\pi} \frac{1}{(5-3z)^2}\frac{dz}{iz}=\frac{1}{i}\int_0^{2\pi}\frac{dz}{z(5-3z)^2}

so we have 2 poles: a simple pole at z=0 and a second order pole at z=5/3

I'm not sure wether i need to include both residues. the question does not indicate the contour to integrate. Should it be a unit circle? I'm not sure why it would be. any ideas?
 
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knowlewj01 said:
I'm not sure wether i need to include both residues. the question does not indicate the contour to integrate. Should it be a unit circle? I'm not sure why it would be. any ideas?

The original integral is over θ from 0 to 2π. When you change your integration variable to z=exp(iθ), what values can z take?
 

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