# Method of residues

## Homework Statement

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

## 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 im 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?

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....

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

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

Are you sure?

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?

not sure...

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.

What is pi - (-pi)?

2$\pi$ thanks :)

^ 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

**** thats not right either

Now i'm getting $\int \frac{-4z^2}{z^4-6z^2+2}\frac{dz}{iz}$ which doesnt work out that easy. So im guessing I got it wrong again. im 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 havent a clue...

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

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