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**[SOLVED] Checking an Integral**

I evaluated the integral below using Cauchy's Formula but I get a different answer than what the book has. The integral is

[tex]\int_0^{2\pi} \frac{d\theta}{a + b\cos \theta}[/tex]

where a > b > 0. Here's what I did: I make the subsitution [itex]z = e^{i\theta}[/itex] so that [itex]a + b\cos\theta = a +b/2(z + z^{-1})[/itex] and [itex]d\theta = dz/(iz)[/itex]. Thus the integral becomes

[tex]\int_{|z|=1} \frac{2 dz}{ibz^2 + i2az +ib}[/tex]

which I can rewrite as

[tex]\int_{|z|=1} \frac{2 dz}{(z - p)(z - q)}[/tex]

where [itex]p = -a/b + \sqrt{b^2 - a^2}/(ib)[/itex] and [itex]q = -a/b - \sqrt{b^2 - a^2}/(ib)[/itex]. Now I can apply Cauchy's Formula by letting f(z) = 2/(z - p) so that the above integral equals [itex]2\pi i f(q) = 2\pi b / \sqrt{b^2 - a^2}[/itex]. Now according to the book, the answer [itex]\pi/\sqrt{a^2 - b^2}[/itex]. I rechecked my algebra and I don't see anything wrong. Is the book wrong?