Complex integration of real-valued trig function

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SUMMARY

The integral of the function \(\int \frac{1}{(3+2\cos(\theta))} d\theta\) evaluated from 0 to \(\pi\) can be computed using complex analysis techniques. The integral is analytic everywhere, leading to the conclusion that its value along any closed loop is zero. A suggested method involves a variable substitution \(t = \theta - \frac{\pi}{2}\), transforming the integral into \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{3 - 2\sin(t)} dt\), which can be further evaluated using the substitution \(u = \tan(t)\) to extend the limits of integration.

PREREQUISITES
  • Understanding of complex analysis and analytic functions
  • Familiarity with trigonometric identities and integrals
  • Knowledge of variable substitution techniques in integration
  • Experience with singularities in integrals
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  • Study the properties of analytic functions and their integrals
  • Learn about variable substitution in definite integrals
  • Explore the method of residues in complex integration
  • Investigate the evaluation of improper integrals with singularities
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Students studying calculus, particularly those focusing on integration techniques, as well as mathematicians and educators looking to deepen their understanding of complex integration methods.

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Homework Statement



Integrate:

[tex]\int \frac{1}{(3+2cos(θ))} dθ[/tex] evaluated from zero to pi.

Homework Equations



I can't think of any. All of the integration formulas in the text rely on the existence of a singularity somewhere in the complex plane. This thing is analytic everywhere.

The Attempt at a Solution



Since it's analytic everywhere, its integral over any closed loop equals zero. That tells me that its integral along the x-axis from zero to pi is equal to the negative of its integral along any other path from pi to zero. I tried to find a path along which cos(theta) is linear (which would give it average value zero) to no avail. I have no idea where to go.
 
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Your integral gives Pi/sqrt(5), calculate the plain ol' indefinite integral and plug in theta=Pi and theta=0.

If you have to do it using complex tricks, I suggest a variable change t = theta-pi/2, which gives [tex]\int_{-\pi/2}^{\pi/2} \frac{1}{3 - 2 \sin t} \mathrm{d} t[/tex], and then you can stretch your integration interval using u = tan t. This gives you an integral from -infty to +infty which has singularities for I am u \neq 0. I'm sure you know how to do this then, if not, just ask again.
 

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