Calculating integrals using residue & cauchy & changing plan

[zhillyz]

Homework Statement

\int_{0}^{2\pi} \dfrac{d\theta}{3+tan^2\theta}

Homework Equations

\oint_C f(z) = 2\pi i \cdot R
R(z_{0}) = \lim_{z\to z_{0}}(z-z_{0})f(z)

The Attempt at a Solution

I did a similar example that had the form
\int_{0}^{2\pi} \dfrac{d\theta}{5+4cos\theta}

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

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

e^{i\theta} = cos(\theta)+isin(\theta)
e^{-i\theta} = cos(\theta)-isin(\theta)
\therefore cos(\theta) = \dfrac{1}{2}\left[e^{i\theta} + e^{-i\theta}\right] = \dfrac{1}{2}\left[z + \dfrac{1}{z}\right]
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 tan^2(\theta) in the same way.

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

Thank you in advance.

 

[Svein]

I do not think the calculus of residues will be of any help here (the integrand has no poles). Some observations:

• 1+\tan^{2}\theta=\frac{1}{\cos^{2}\theta}
• \cos(2\theta)=2\cos^{2}\theta-1, so \cos^{2}\theta=\frac{1+\cos2\theta}{2}

 

[vela]

I do not think the calculus of residues will be of any help here (the integrand has no poles).

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

I get kind of lost though trying to describe tan^2(\theta) in the same way.

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.