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**[SOLVED] Integral of Square of Reciprocal of √(a + cos t)**

**Homework Statement**

Demonstrate that

[tex]\int_0^\pi \frac{dt}{(a + \cos t)^2} = \frac{a\pi}{(\sqrt{a^2 - 1})^3}[/tex]

where a > 0.

**The attempt at a solution**

I want to compute this using the Residue Theorem. The first thing I did was make the substitution [itex]\theta / 2 = t[/itex] so that

[tex]\int_0^\pi \frac{dt}{(a + \cos t)^2} = \int_0^{2\pi} \frac{d\theta}{2(a + \cos \theta/2)^2}[/tex]

I then made the substitution [itex]z = e^{i\theta/2}[/itex] so that

[tex]\int_0^{2\pi} \frac{d\theta}{2(a + \cos \theta/2)^2} = \int_{|z|=1} \frac{4z}{i(z^2 + 2az + 1)^2} \, dz = \int_{|z|=1} \frac{4z}{i(z - z_0)^2(z - z_1)^2} \, dz[/tex]

where

[tex]\begin{align*}z_0 &= \frac{-a + \sqrt{a^2 - 1}}{2}\\

z_1 &= \frac{-a - \sqrt{a^2 - 1}}{2}\end{align*}[/tex]

Call the integrand of the last integral f(z). By the Residue Theorem

[tex]\int_{C} f(z) \, dz = 2\pi i \sum_{u \in D} \text{Res}(f; u)[/tex]

where u is a pole of f, C is the unit circle and D is its interior. The only possible poles are [itex]z_0, z_1[/itex]. I know [itex]z_0[/itex] is in D for all a > 0. However, [itex]z_1[/itex] is not in D if a > 5/4. What do I in this case? I would condition on the value of a but then my answer won't match up with that of the problem statement.