Simple residue theorem question.

[QuantumLuck]

Homework Statement

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).

Homework Equations

Given integral

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.

 

[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.

 

[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).

 

[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).

 

[QuantumLuck]

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

 

[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.