- #1
zhillyz
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Homework Statement
[tex]\int_{0}^{2\pi} \dfrac{d\theta}{3+tan^2\theta}[/tex]
Homework 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]
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.