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Solving Trigonometric integrals using cauchy residue theorem

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    evaluate the given trigonometric integral

    ∫1/(cos(θ)+2sin(θ)+3) dθ

    where the lower limit is 0 and the upper limit is 2π

    2. Relevant equations

    z = e^(iθ)
    cosθ = (z+(z)^-1)/2
    sinθ = (z-(z)^-1)/2i
    dθ = dz/iz

    3. The attempt at a solution

    after I substitute and simplify the integral i get this

    2∫1/[(2+i)z^2 + 6iz + (i-2)] dz

    This is being integrated over a closed surface. The closed surface is a the unit circle (aka |z| = 1)

    when I set the denominator to zero and solve for the poles I end up getting poles at
    z1 = -i/(2+i) and z2 = -5i/(2+i)

    this is where i get confused. When i start to do the residue theorem i use the residue at
    z1 = -i/(2+i) because z1 lies within the unit circle and z2 does not. The solution that i have uses both the residues at z1 and z2. If anyone can give me some insight on why i have to use both residues that would be great. The only thing i can think of is that i am not integrating over the unit circle but that would not make sense because i have parameterized the original integrand into a contour integral where the contour is the unit circle.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 31, 2012 #2
    You have to compute the residue only at the pole inside the unit disk.The other pole is irrelevant for this integral.
     
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