Evaluate real integral using residue theorem, where did I go wrong?

[Poopsilon]

Edit: Never mind I found my error, moderator can lock this.

Homework Statement

Evaluate the integral $\int_0^{\pi} \frac{dt}{(a+cost)^2}$ for a > 1.

Homework Equations

$$\int_0^{\pi}\frac{dt}{(a+cost)^2} = \pi i\sum_{a\epsilon \mathbb{E}}Res(f;\alpha)$$

Where $\mathbb{E}$ is the open unit disk, and $f(z) = \frac{1}{iz(a+\frac{1}{2}(z+\frac{1}{z}))^2}$.

The Attempt at a Solution

$$f(z) = \frac{1}{iz(a+\frac{1}{2}(z+\frac{1}{z}))^2} = \frac{1}{iz\frac{(z^2 + 2az + 1)^2}{4z^2}} = \frac{-4iz}{(z^2 + 2az + 1)^2} = \frac{-4iz}{[(z + a + \sqrt{a^2 - 1})(z + a - \sqrt{a^2 - 1})]^2}$$

Thus f has two roots of multiplicity two. It's fairly easy to see that only the root $z = -a + \sqrt{a^2 - 1}$ lies inside the open unit disk for a > 1, thus we set $b = -a + \sqrt{a^2 - 1}$ and obtain:

$$\pi i\sum_{\alpha\epsilon\mathbb{E}}Res(f;\alpha) = Res(f;b) = lim_{z\rightarrow b}\frac{d}{dz}\frac{-4iz(z+a-\sqrt{a^2 - 1})^2}{(z + a + \sqrt{a^2 - 1})^2(z + a - \sqrt{a^2 - 1})^2} = lim_{z\rightarrow b}4\pi\frac{d}{dz}\frac{z}{(z+a+\sqrt{a^2 - 1})^2} = lim_{z\rightarrow b}4\pi \frac{a + \sqrt{a^2 - 1} -z}{(z + a + \sqrt{a^2 - 1})^3}$$

From here I take the limit and after simplification obtain $\frac{\pi\sqrt{a^2 - 1}}{a^3}$. Yet the book says my answer should be $\frac{\pi a}{(a^2 - 1)\sqrt{a^2 - 1}}$. Where have I gone wrong? Thanks.

 

[Alpha Floor]

Explain your mistake so that everyone can learn from it!

 

[Poopsilon]

It wasn't really a mistake in my understanding of the theory it was just a sign error. I was accidentally plugging in -b instead of b.