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Using Residue Calculus to evaluate real integrals

  1. Jul 4, 2010 #1
    I need to calculate the definite integral of dx/(4+3cos(x)) from 0 to 2pi.

    I believe that the integral from 0 to 2pi of (f(e^ix))*i*(e^ix)dx is equivalent to the integral over the unit circle of f(z)dz.
    If that's true, then this problem boils down to finding f(z) so that f(z)iz=1/(4+(3/2)(z+z^-1) since cosx=(1/2)(e^ix+e^-1x).
    This gives f(z)=-2i/(3z^2+8z+3) which has poles at (-4-√7)/3 and (-4+√7)/3, and only the latter lies within the unit circle, so the value of our integral ought to just be 2*pi*i*Res(f,(-4+√7)/3)=(2*pi*i)*1/((-4+√7)/3-(-4-√7)/3)=2*pi*i*(3/2√7)=6*pi/√7.

    However, when I checked my result on Wolframalpha, I was off by a factor of 3. They had 2*pi/√7.

    Where did I go wrong?

    Note: I also tried to calculate thedefinite integral of dx/(5+4sin(x)) from 0 to 2pi and ended up off by a factor of 2 according again to Wolfram.

    Please help.
  2. jcsd
  3. Jul 4, 2010 #2
    Nevermind. I found it. I had forgotten about the coefficient on z^2 when I wrote the denominator in factored form which brings an additional factor of 3 into the denominator.
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