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Calculating integrals using residue & cauchy & changing plan

  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_{0}^{2\pi} \dfrac{d\theta}{3+tan^2\theta}[/tex]

    2. Relevant equations
    [tex]\oint_C f(z) = 2\pi i \cdot R[/tex]
    [tex]R(z_{0}) = \lim_{z\to z_{0}}(z-z_{0})f(z)[/tex]

    3. The attempt at a solution
    I did a similar example that had the form
    [tex]\int_{0}^{2\pi} \dfrac{d\theta}{5+4cos\theta}[/tex]

    where I would change to the complex plane [itex]z[/itex] where [itex]z = e^{i\theta}[/itex] and so [itex] dz = ie^{i\theta}d\theta \to d\theta = \dfrac{e^{-i\theta}dz}{i} [/itex]

    The cosine function could also be written in terms of the exponential function as such;

    [tex] e^{i\theta} = cos(\theta)+isin(\theta)[/tex]
    [tex]e^{-i\theta} = cos(\theta)-isin(\theta)[/tex]
    [tex]\therefore cos(\theta) = \dfrac{1}{2}\left[e^{i\theta} + e^{-i\theta}\right] = \dfrac{1}{2}\left[z + \dfrac{1}{z}\right][/tex]
    after you substitute all these back into the formula it gives a denominator that can be factorised to give poles which you can find the residual values for and then calculate the integral using the residue theorem. I get kind of lost though trying to describe [itex] tan^2(\theta)[/itex] in the same way.

    [tex] tan^2(\theta) = sec^2(\theta) - 1 = \dfrac{sin^2(\theta)}{cos^2(\theta)}[/tex]
    [tex] sin^2(\theta) + cos^2(\theta) = 1[/tex]

    Thank you in advance.
     
  2. jcsd
  3. Oct 25, 2015 #2

    Svein

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    I do not think the calculus of residues will be of any help here (the integrand has no poles). Some observations:
    • [itex]1+\tan^{2}\theta=\frac{1}{\cos^{2}\theta} [/itex]
    • [itex]\cos(2\theta)=2\cos^{2}\theta-1 [/itex], so [itex] \cos^{2}\theta=\frac{1+\cos2\theta}{2}[/itex]
     
  4. Oct 25, 2015 #3

    vela

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    I think you did something wrong. There are three poles inside the unit circle.

    You can show that
    $$\sin \theta = \frac{e^{i\theta}-e^{-i\theta}}{2i} = \frac{z-\frac 1z}{2i},$$ so
    $$\tan \theta = \frac 1i \frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}} = \frac 1i \frac{z^2-1}{z^2+1}.$$ Then it's just a matter of doing the algebra.
     
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