Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x

Theorem

 * $\displaystyle \int_0^{2 \pi} \frac {\d x} {a + b \cos x} = \frac {2 \pi} {\sqrt {a^2 - b^2} }$

where $a$ and $b$ are real numbers with $a > b > 0$.

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 where:


 * $\displaystyle \paren {z + \frac a b}^2 - \frac {a^2} {b^2} + 1 = 0$

That is, where:


 * $\displaystyle \size {z + \frac a b} = \frac {\sqrt {a^2 - b^2} } b$

So:


 * $z_1 = \dfrac {-a + \sqrt {a^2 - b^2} } b$

and:


 * $z_2 = \dfrac {-a - \sqrt {a^2 - b^2} } b$

are the poles of the integrand.

We have:

So $z_2$ lies outside the closed disk $\size z \le 1$ for all real $a > b > 0$, so is of no concern.

We now establish a bound on $z_1$.

As $a > b > 0$, we have that:


 * $a^2 > a^2 - b^2 > 0$

So, from Square Root is Strictly Increasing:


 * $a > \sqrt {a^2 - b^2} > 0$

Multiplying through $-2 \sqrt {a^2 - b^2} < 0$:


 * $-2 a \sqrt{a^2 - b^2} < 2 \paren {b^2 - a^2}$

This can be rewritten as:

giving:


 * $\size {-a + \sqrt {a^2 - b^2} } < b$

Therefore:


 * $\size {z_1} = \size {\dfrac {-a + \sqrt {a^2 - b^2} } b} < 1$

So $z_1$ is the only pole of the integrand lying within $C$.

Therefore:

Also see

 * Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \sin x}$