Definite Integral from 0 to 2 Pi of Reciprocal of One minus 2 a Cosine x plus a Squared

Theorem

 * $\ds \int_0^{2 \pi} \frac {\d x} {1 - 2 a \cos x + a^2} = \frac {2 \pi} {1 - a^2}$

where $a$ is a real number with $0 < a < 1$.

Proof
Let $C$ be the unit open disk centred at $0$.

The boundary of $C$, $\partial C$, can be parameterized by:


 * $\map \gamma \theta = e^{i \theta}$

for $0 \le \theta \le 2 \pi$.

We have:

The integrand has poles:


 * $z_1 = a$

and:


 * $z_2 = \dfrac 1 a$

We have $0 < a < 1$, so $\dfrac 1 a > 1$.

So $z_2$ lies outside the closed disk $\size z \le 1$, and $z_1$ lies in $C$, so the only pole of concern is $z_1$.

Therefore: