1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted