# Computing a residue

To find the value of $\int_0^\pi \frac{\sin^2 \theta}{a + \cos \theta} \, d\theta$ with $a > 1$ using residue theory, I am trying to compute the residue at the simple pole $z=-a+\sqrt{a^2-1}$ inside the unit disc. I tried both ways: \begin{align} \operatorname{Res}\left(\frac{z^4-2z^2+1}{z^4+2az^3+z^2},-a+\sqrt{a^2-1}\right) &= \frac{z^4-2z^2+1}{\frac d{dz}(z^4+2az^3+z^2)}\Bigg\vert_{z=-a+\sqrt{a^2-1}} \\ &= \frac{z^4-2z^2+1}{4z^3+6az+2z}\Bigg\vert_{z=-a+\sqrt{a^2-1}} \end{align} and \begin{align} \operatorname{Res}\left(\frac{z^4-2z^2+1}{z^4+2az^3+z^2},-a+\sqrt{a^2-1}\right) &= \lim_{z \to -a+\sqrt{a^2-1}} (z-(-a+\sqrt{a^2-1}))\frac{z^4-2z^2+1}{z^4+2az^3+z^2} \\ &= \lim_{z \to -a+\sqrt{a^2-1}} \frac{z^4-2z^2+1}{z^2(z-(-a-\sqrt{a^2-1}))}. \end{align} Is there a clean way to finding this particular residue? Because the methods I tried above require chugging through long arithmetic calculations.

• Your root is a zero of $z^2+2 a z+1$, so you only need to take the derivative of that term. Jul 13 '18 at 20:08
• Okay, that helps ease up the claculation in the denominator. What about the numerator? Should I break the fraction apart and simplify the first two terms of the three resulting fractions? Jul 13 '18 at 20:10
• It doesn't matter. Jul 13 '18 at 20:12

\begin{align} \frac{\sin^2(\theta)}{a+\cos(\theta)}=\frac{1-a^2}{a+\cos(\theta)}+a-\cos(\theta) \end{align}
The integral of the term $a-\cos(\theta)$ is trivial. If we wish to use contour integration for the term $\displaystyle \frac{1-a^2}{a+\cos(\theta)}$, we can proceed to write for $a>1$
\begin{align} \int_0^\pi \frac{1-a^2}{a+\cos(\theta)}\,d\theta&=i(a^2-1) \oint_{|z|=1} \frac{1}{z^2+2az+1}\,dz\\\\ &=2\pi (1-a^2)\text{Res}\left(\frac{1}{z^2+2az+1}, z=-a+\sqrt{a^2-1}\right)\\\\ &=-\pi\sqrt{a^2-1} \end{align}
$$\int_0^\pi \frac{\sin^2(\theta)}{a+\cos(\theta)}\,d\theta=\pi(a-\sqrt{a^2-1})$$