# Homework Help: Evaluate Definite Integral with Complex Analysis

1. Sep 28, 2014

### KleZMeR

1. The problem statement, all variables and given/known data

$I_1 = \int_0^{2\pi} \frac{sin\theta}{3+2cos\theta} d\theta$

2. Relevant equations

Using identities to change from cos, sin, to variables of z, I get:

$2iz^2 + 6iz + 2i$ in my denominator

3. The attempt at a solution

Looking for a singularity, will I use a quadratic for this denominator to find my two zeros and then use my positive result to calculate the residue?

2. Sep 28, 2014

### KleZMeR

Also, is this a pole of second order? Yes, I think it is. But what this means is that I will have to take a derivative of this fraction with polynomials? I have tried to simplify it into products but to no avail.

3. Sep 28, 2014

### vela

Staff Emeritus
You forgot to write $d\theta$ in terms of $dz$.

Yes, you need to use the quadratic equation to factor the denominator. All of the poles in this problem are simple poles.

4. Sep 28, 2014

### KleZMeR

Thanks Vela, I had that written but didn't include that. So, I found my residue for my singularity at $z_0 = 0$
which was $-\frac{1}{2}$

But when I find my residue for the $z_0 = \frac{\sqrt5+3}{2}$ (the value I got from my quadratic)
I got a residue = 0.
Quadratic being $(z^2)+3z+1$

What do you think? I see from some examples I've done that this IS possible, but this is homework so I'd like to know what you think?

5. Sep 28, 2014

### vela

Staff Emeritus
Show your calculation of the residue. It shouldn't be 0. The integral, however, turns out to be equal to 0. You can see this by changing the interval of integration to $-\pi$ to $\pi$ and noting the integrand is an odd function.

6. Sep 28, 2014

### KleZMeR

Well for starters I am not getting my ($z-z_0$) to cancel out, should I be using exponents instead of my z identities for the trig functions? The only reason I was able to get my residue at 0 was because $\frac{z-0}{z}$ occurred.

7. Sep 28, 2014

### vela

Staff Emeritus
From what you just said, it's not clear at all to me what you're doing to evaluate the residue.

8. Sep 28, 2014

### KleZMeR

Using $cos\theta = \frac{1}{2}(z+\frac{1}{z})$ and $sin\theta = -\frac{1}{2i}(z-\frac{1}{z})$

my residue simplifies to $-\frac{(z-z_0)(z^2-1)}{z(6z+2z^2+2)}$

At $z_0 = 0$

I get

$Res[f,0] = 1/2$

but my other two roots, of the polynomial in the denominator, are giving me problems when trying to compute the residue.

$\frac{1}{2}(\sqrt5-3)$ and $\frac{1}{2}(-3-\sqrt5)$

I think both these points are in my region? it's a full revolution yes?

9. Sep 28, 2014

### KleZMeR

I'm trying LHopital

10. Sep 29, 2014

### KleZMeR

Seemed to have worked! With my root that is within the circle $\frac{1}{2}(\sqrt5-3)$ I get $\frac{1}{2}$

And so $\frac{1}{2} + (-\frac{1}{2}) = 0 = Res[f,0] + Res[f,\frac{1}{2}(\sqrt5-3)]$

11. Sep 29, 2014

### vela

Staff Emeritus
I wouldn't call that the residue. You mean the residue for a simple pole at $z=z_0$ is given by
$$b_1 = \lim_{z \to z_0} -\frac{(z-z_0)(z^2-1)}{z(6z+2z^2+2)}.$$

I still don't know what problem you ran into.