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

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

Proof:
Let $C$ be the unit open disk in the complex plane centred at $0$. 
The boundary of $C$, $\partial C$, can be parameterized by: 
:$\map z x = e^{i x}$
for $0 \le x \le 2 \pi$. 
We have:
{{begin-eqn}}
{{eqn  | l = \int_0^{2 \pi} \frac {\d x} {a + b \cos x}
       | r = \int_0^{2 \pi} \frac {\d x} {a + \frac b 2 \paren {e^{i x} + e^{-i x} } }
       | c = Euler's Cosine Identity
}}
{{eqn  | r = \int_0^{2 \pi} \paren {\frac {2 e^{i x} } {2 e^{i x} } } \frac {\d x} {a + \frac b 2 \paren {e^{i x} + e^{-i x} } }
       | c = multiplying by $1$
}}
{{eqn  | r = 2 \int_0^{2 \pi} \frac {e^{i x} \rd x } {2 a e^{i x} + b e^{2 i x} + b} 
       | c =
}}
{{eqn  | r = 2 \int_0^{2 \pi} \frac {e^{i x} \rd x } {b \paren {e^{2 i x} + \frac {2 a} b e^{i x} + 1} } 
       | c = rearranging
}}
{{eqn  | r = \frac 2 {i b} \int_{\partial C} \frac {\d z} {z^2 + \frac {2 a} b z + 1}
       | c = {{Defof|Complex Contour Integral}}, Derivative of Exponential Function, $\rd z = i e^{i x} \rd x$
}}
{{eqn  | r = \frac 2 {i b} \int_{\partial C} \frac {\d z} {\paren {z + \frac a b}^2 - \frac {a^2} {b^2} + 1}
       | c = Completing the Square
}}
{{end-eqn}}
The integrand has poles where: 
:$\paren {z + \dfrac a b}^2 - \dfrac {a^2} {b^2} + 1 = 0$
That is, where: 
:$\size {z + \dfrac a b} = \dfrac {\sqrt {a^2 - b^2} } b$
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: 
{{begin-eqn}}
{{eqn | l = \size {z_2}
      | r = \size {\frac {-a - \sqrt {a^2 - b^2} } b}
}}
{{eqn | r = \frac {a + \sqrt {a^2 - b^2} } b
      | c = as $a > b > 0$
}}
{{eqn | r = \frac a b \paren {1 + \sqrt {1 - \frac {b^2} {a^2} } }
}}
{{eqn | o = >
      | r = \frac a b
      | c = as $\sqrt {1 - \dfrac {b^2} {a^2} } > 0$
}}
{{eqn | o = >
      | r = 1
}}
{{end-eqn}} 
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 by $-2 \sqrt {a^2 - b^2}$: 
:$-2 a \sqrt{a^2 - b^2} < 2 \paren {b^2 - a^2} < 0$
This can be rewritten as: 
{{begin-eqn}}
{{eqn | l = a^2 - 2 a \sqrt {a^2 - b^2} + a^2 - b^2 
      | r = a^2 - 2 a \sqrt {a^2 - b^2} + \paren {\sqrt {a^2 - b^2} }^2
}}
{{eqn | r = \paren {-a + \sqrt {a^2 - b^2} }^2
      | c = Square of Sum
}}
{{eqn | o = <
      | r = b^2
}}
{{end-eqn}}
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: 
{{begin-eqn}}
{{eqn  | l = \frac 2 {i b} \int_C \frac {\d z} {z^2 + \frac {2 a} z + 1}
       | r = \frac {4 \pi i} {i b} \Res {\frac 1 {z^2 + \frac {2 a} b z + 1} } {z_1}
       | c = Cauchy's Residue Theorem
}}
{{eqn  | r = \frac {4 \pi} b \paren {\lim_{z \mathop \to {z_1} } \paren {z - z_1} \frac 1 {z^2 + \frac {2 a} b z + 1} }
       | c = Residue at Simple Pole
}}
{{eqn  | r = \frac {4 \pi} b \paren {\lim_{z \mathop \to {\frac {-a + \sqrt {a^2 - b^2} } b} } \paren {z - \frac {-a + \sqrt {a^2 - b^2} } b} \frac 1 {\paren {z - \frac {-a + \sqrt {a^2 - b^2} } b} \paren {z - \frac {-a - \sqrt {a^2 - b^2} } b} } }
       | c = 
}}
{{eqn  | r = \frac {4 \pi} b \sqbrk {\frac 1 {\paren {z - \frac {-a - \sqrt {a^2 - b^2} } b} } }_{z = \frac {-a + \sqrt {a^2 - b^2} } b}
       | c = cancelling $\paren {z - \dfrac {-a + \sqrt {a^2 - b^2} } b} $
}}
{{eqn  | r = \frac {4 \pi} b \paren { {\frac 1 {\paren {\frac {-a + \sqrt {a^2 - b^2} } b - \frac {-a - \sqrt {a^2 - b^2} } b} } } }
       | c = 
}}
{{eqn  | r = \frac {4 \pi} b \paren { {\frac b {2 \paren {\sqrt {a^2 - b^2} } } } }
       | c = 
}}
{{eqn  | r = \frac {2 \pi} {\sqrt {a^2 - b^2} }
}}
{{end-eqn}}
{{qed}}