10. $[8]$ Let $a$ and $b$ be real numbers satisfying $a>b>0$. Evaluate
$$
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta \text {. }
$$

Express your answer in terms of $a$ and $b$.
Solution: Using the geometric series formula, we can expand the integral as follows:
$$
\begin{aligned}
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta & =\frac{1}{a} \int_{0}^{2 \pi} 1+\frac{b}{a} \cos (\theta)+\left(\frac{b}{a}\right)^{2} \cos ^{2}(\theta) d \theta \\
& =\frac{1}{a} \sum_{n=0}^{\infty} \int_{0}^{2 \pi}\left(\frac{b}{a}\right)^{n}\left(\frac{e^{i \theta}+e^{-i \theta}}{2}\right)^{n} d \theta \\
& =\frac{2 \pi}{a} \sum_{n=0}^{\infty}\left(\frac{b^{2}}{a^{2}}\right)^{n} \frac{\binom{2 n}{n}}{2^{2 n}} d \theta
\end{aligned}
$$

To evaluate this sum, recall that $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$th Catalan number. The generating function for the Catalan numbers is
$$
\sum_{n=0}^{\infty} C_{n} x^{n}=\frac{1-\sqrt{1-4 x}}{2 x}
$$
and taking the derivative of $x$ times this generating function yields $\sum\binom{2 n}{n} x^{n}=\frac{1}{\sqrt{1-4 x}}$. Thus the integral evaluates to $\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}$, as desired.