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## Homework Statement

evaluate using the method of residues [itex]\int^{\pi}_{-\pi}[/itex] [itex]\frac{d\theta}{1+sin^2\theta}[/itex] (=[itex]\pi\sqrt{2})[/itex]

## Homework Equations

## The Attempt at a Solution

[itex] sin^2 \theta = \frac{1}{2i}(z^2 - \frac{1}{z^2}) [/itex] so our integral becomes [itex]\int \frac{2iz^2}{z^4+2iz^2-1}\frac{dz}{iz} [/itex]but before i continue on from here im not sure what the story with the limits. Normally i've just seen limits of 0 to 2[itex]\pi[/itex] so should iI put a [itex]\frac{1}{2}[/itex] in from of the integral?