Method of Residues: Evaluating \int_{-\pi}^{\pi}\frac{d\theta}{1+sin^2\theta}

[gtfitzpatrick]

Homework Statement

evaluate using the method of residues $\int^{\pi}_{-\pi}$ $\frac{d\theta}{1+sin^2\theta}$ (=$\pi\sqrt{2})$

Homework Equations

The Attempt at a Solution

$sin^2 \theta = \frac{1}{2i}(z^2 - \frac{1}{z^2})$ so our integral becomes $\int \frac{2iz^2}{z^4+2iz^2-1}\frac{dz}{iz}$

but before i continue on from here I am not sure what the story with the limits. Normally I've just seen limits of 0 to 2$\pi$ so should iI put a $\frac{1}{2}$ in from of the integral?

 

[gtfitzpatrick]

sorry got that wrong $sin^2 \theta = (\frac{1}{2i}(z - \frac{1}{z}))^2$ should be $\int \frac{-4z^2}{z^4-2z^2+1}\frac{dz}{iz}$ but still unsure about my limits...

 

[awkward]

As theta varies from -pi to pi, how much of the unit circle does it sweep out?

 

[gtfitzpatrick]

I think its Half a sweep, that's why i think i put a half in front of the integral

 

[awkward]

Are you sure?

 

[gtfitzpatrick]

a whole sweep is 2$\pi$, half a sweep is o to $\pi$ so is $\pi$ to -$\pi$ half a sweep...backwards? so i put -$\frac{1}{2}$ in front of it?

 

[gtfitzpatrick]

not sure...

 

[awkward]

What is pi - (-pi)?

 

[uart]

sorry got that wrong $sin^2 \theta = (\frac{1}{2i}(z - \frac{1}{z}))^2$ should be $\int \frac{-4z^2}{z^4-2z^2+1}\frac{dz}{iz}$ but still unsure about my limits...

^ Also, double check your denominator there, you've made a mistake in the algebra.

When you sort that out you'll find it comes out fairly easy, four real roots with two of them inside your contour.

 

[gtfitzpatrick]

What is pi - (-pi)?

2$\pi$ thanks :)

 

[gtfitzpatrick]

^ Also, double check your denominator there, you've made a mistake in the algebra.

When you sort that out you'll find it comes out fairly easy, four real roots with two of them inside your contour.

$\int \frac{-4z^2}{z^4-2z^2+2}\frac{dz}{iz}$

thanks a million

 

[gtfitzpatrick]

**** that's not right either

 

[gtfitzpatrick]

Now I'm getting $\int \frac{-4z^2}{z^4-6z^2+2}\frac{dz}{iz}$ which doesn't work out that easy. So I am guessing I got it wrong again. I am going to use formula and see what it throws up anywho.thanks guys. Do either of you have a clue what my other post (jordans lemma) is about? i haven't a clue...

 

[awkward]

I think you need to double-check your algebra, one more time.

 

[gtfitzpatrick]

yes your right, it should be $\int \frac{-4z^2}{z^4-6z^2+1}\frac{dz}{iz}$