Example 14. Calculate the integral

$$
I=\int_{0}^{2 \pi} \frac{d x}{(a+b \cos x)^{2}} \quad(a>b>0)
$$
Solution. Applying the substitution $e^{i z}=z$, we obtain after simple transformations

$$
I=\frac{4}{i} \int_{C} \frac{z d z}{\left(b z^{2}+2 a z+b\right)^{2}}=\frac{4}{i} 2 \pi i \sum_{k=1}^{n} \operatorname{res} F\left(z_{k}\right)
$$

Inside the unit circle, under the condition that $a>b>0$, there is only one zero (double)

$$
z_{1}=\frac{-a+\sqrt{a^{2}-b^{2}}}{b}
$$

The residue of the function

$$
F(z)=\frac{z}{\left(b z^{2}+2 a z+b\right)^{2}}
$$

with respect to this pole

$$
\operatorname{res} F\left(z_{1}\right)=\lim _{x \rightarrow z_{1}} \frac{d}{d z}\left(\frac{z\left(z-z_{1}\right)^{2}}{b^{2}\left(z-z_{1}\right)^{2}\left(z-z_{2}\right)^{2}}\right)=\frac{a}{4}\left(a^{2}-b^{2}\right)^{-3 / 2}
$$

Thus,

$$
I=\frac{2 \pi a}{\left(a^{2}-b^{2}\right)^{3 / 2}}
$$

## Problems for Independent Solution

Calculate the following integrals:

409. $\int_{0}^{2 \pi} \frac{d x}{1-2 p \cos x+p^{2}}(01)$.
411. $\int_{0}^{2 \pi} \frac{\cos x d x}{1-2 p \sin x+p^{2}} \quad(01)$.
413. $\int_{0}^{\pi} \operatorname{ctg}(x-a) d x \quad(\operatorname{Im} a>0)$.
414. $\int_{0}^{2 \pi} \frac{\sin ^{2} x}{a+b \cos x} d x \quad(a>b>0)$.
415. $\int_{0}^{2 \pi} \frac{d x}{1+a \cos x} \quad(0<a<1)$.
416. $\int_{0}^{2 \pi} \frac{d x}{a+b \cos x}(0<b<a)$.

$3^{\circ}$. Summation of some series using residues. 1. Let the function $f(z)$ be analytic on the entire complex plane except for a finite number of poles $z_{1}, z_{2}, \ldots, z_{k}$, none of which coincide with any of the points $z=0, \pm 1, \pm 2, \ldots$, and let $f(z)$ satisfy the condition
$f(z)=O\left(z^{-2}\right)$ as $z \rightarrow \infty^{1)}$. Then

$$
\sum_{n=-\infty}^{\infty} f(n)=-\pi \sum_{m=1}^{k} \operatorname{res}_{z=z_{m}}[f(z) \operatorname{ctg} \pi z]
$$