Solving Trigonometric integrals using cauchy residue theorem

In summary, to evaluate the given trigonometric integral ∫1/(cos(θ)+2sin(θ)+3) dθ from 0 to 2π, the residue theorem is used to find the residue at the pole -i/(2+i) since it lies within the unit circle. The other pole, -5i/(2+i), is not used in the calculation. The integral is being integrated over a closed surface, specifically the unit circle. The integrand is also parameterized into a contour integral with the contour being the unit circle.
  • #1
furth721
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Homework Statement



evaluate the given trigonometric integral

∫1/(cos(θ)+2sin(θ)+3) dθ

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

Homework Equations



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

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.
 
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  • #2
You have to compute the residue only at the pole inside the unit disk.The other pole is irrelevant for this integral.
 

FAQ: Solving Trigonometric integrals using cauchy residue theorem

1. How does the Cauchy residue theorem help in solving trigonometric integrals?

The Cauchy residue theorem is a powerful tool in complex analysis that states that the value of a complex integral around a closed path is equal to the sum of the residues of all the singularities inside the path. This theorem can be applied to solve trigonometric integrals by converting them into complex integrals and then using the residue theorem to evaluate them.

2. What are the steps involved in solving a trigonometric integral using the Cauchy residue theorem?

The steps involved in solving a trigonometric integral using the Cauchy residue theorem are as follows:

  • Convert the trigonometric integral into a complex integral by using the Euler's formula.
  • Identify the singularities of the complex function.
  • Find the residues of the singularities inside the contour.
  • Apply the Cauchy residue theorem to evaluate the complex integral.
  • Convert the complex integral back to its original form to get the solution of the trigonometric integral.

3. Can the Cauchy residue theorem be applied to all types of trigonometric integrals?

No, the Cauchy residue theorem can only be applied to integrals that can be converted into complex integrals and have singularities inside the contour. Trigonometric integrals that do not have singularities or have singularities outside the contour cannot be solved using this theorem.

4. What are the advantages of using the Cauchy residue theorem to solve trigonometric integrals?

The Cauchy residue theorem provides a systematic and efficient way to solve trigonometric integrals. It reduces the complexity of the integral and makes it easier to evaluate. It also allows for the use of complex analysis techniques to solve the integral, which may not be possible with traditional methods.

5. Are there any limitations to using the Cauchy residue theorem to solve trigonometric integrals?

Yes, there are some limitations to using the Cauchy residue theorem. It can only be applied to integrals that can be converted into complex integrals and have singularities inside the contour. Also, finding the residues of the singularities can be a challenging task in some cases. Additionally, the contour used in the theorem must be carefully chosen to ensure that all the singularities are enclosed and the integral is well-defined.

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