# Solving Trigonometric integrals using cauchy residue theorem

1. Oct 30, 2012

### furth721

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. Oct 31, 2012

### hedipaldi

You have to compute the residue only at the pole inside the unit disk.The other pole is irrelevant for this integral.