- #1

MissMoneypenny

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## Homework Statement

Compute the real integral

[itex]\int\frac{dθ}{2+sin(θ)}[/itex], where the limits of integration are from 0 to 2π

by writing the sine function in terms of the exponential function and making the substitution z=e[itex]^{iθ}[/itex] to turn the real integral into a complex integral.

## Homework Equations

sin(z)=[itex]\frac{e^{iz}-e^{-iz}}{2i}[/itex]

Cauchy's integral formula.

## The Attempt at a Solution

I wrote sin(θ)=[itex]\frac{e^{iθ}-e^{-iθ}}{2i}[/itex] and plugged that into the integral. The integrand then simplifies to [itex]\frac{2i}{4i+e^{iθ}+e^{-iθ}}[/itex]. Letting z=e[itex]^{iθ}[/itex] gives dθ=[itex]\frac{dz}{iz}[/itex]. However, plugging this substitution back into the integral gives

[itex]\int\frac{2dz}{-4z+iz^{2}-1}[/itex]

That's where I get stuck. I'm not sure where to go from here. I thought about using partial fractions, but that seems like it may be a nastier approach than necessary. Any help will be much appreciated.

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