# Simple residue theorem question.

1. Mar 10, 2009

### QuantumLuck

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

So I know how to evaluate the integral from 0 to 2pi of 1/2+cos theta. However, the question I am being asked to do has me calculate this integral from 0 to pi. I am not sure what adjustment is necessary to get the integral i am given (from 0 to pi) to the form I know how to calculate (0 to 2pi).

2. Relevant equations
Given integral

3. The attempt at a solution
at first I thought that since cosine is an even function I could merely double the integral's bounds and divide it by two since the integral would just pick up an extra half of a periodic function. but the constant and the fact that cosine is in the denominator made me question that idea. so as of right now i am a bit stuck.

2. Mar 10, 2009

### yyat

If the variable x only appears as cos(x) , then the function is even.
Maybe it would be helpful if you could state what the integrand is.

3. Mar 10, 2009

### QuantumLuck

it is exactly as i said. the integral from 0 to pi of 1/2+cos(theta) dtheta. now what i know how to evaluate is the integral from 0 to 2 pi of the previous integrand. i am just not sure how to modify the equation (if that is the correct path to take anyways).

4. Mar 10, 2009

### yyat

I was confused because you said the cosine was in the denominator.
The most straightforward way to compute the integral is to find the indefinite integral F (aka antiderivative) first, and then calculate F(pi)-F(0).

5. Mar 10, 2009

### QuantumLuck

argh. i am a fool. i forgot the parentheses; cosine is in the denominator. the integrand is dtheta/(2 + cos(theta))

6. Mar 10, 2009

### yyat

The function you are trying to integrate is even and 2pi-periodic, so the integral from 0 to pi is half of the integral from 0 to 2pi, as you guessed.