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Simple residue theorem question.

  1. Mar 10, 2009 #1
    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. jcsd
  3. Mar 10, 2009 #2
    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.
  4. Mar 10, 2009 #3
    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).
  5. Mar 10, 2009 #4
    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).
  6. Mar 10, 2009 #5
    argh. i am a fool. i forgot the parentheses; cosine is in the denominator. the integrand is dtheta/(2 + cos(theta))
  7. Mar 10, 2009 #6
    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.
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