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

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

Proof:
Fix $b \in \R$ and define: 
:$\ds \map I a = \int_0^\pi \map \ln {a + b \cos x} \rd x$
for $a \ge \size b$.
We have: 
{{begin-eqn}}
{{eqn | l = \map {I'} a
      | r = \frac \d {\d a} \int_0^\pi \map \ln {a + b \cos x} \rd x
}}
{{eqn | r = \int_0^\pi \frac \partial {\partial a} \paren {\map \ln {a + b \cos x} } \rd x
      | c = Definite Integral of Partial Derivative
}}
{{eqn | r = \int_0^\pi \frac 1 {a + b \cos x} \rd x
      | c = Derivative of Natural Logarithm
}}
{{eqn | r = \frac 1 2 \int_0^{2 \pi} \frac 1 {a + b \cos x} \rd x
      | c = Definite Integral of Even Function
}}
{{eqn | r = \frac \pi {\sqrt {a^2 - b^2} }
      | c = Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \cos x}$
}}
{{end-eqn}}
So, by Primitive of $\sqrt {x^2 - a^2}$: Logarithm Form: 
:$\map I a = \pi \map \ln {a + \sqrt {a^2 - b^2} } + C$
for all $a \ge \size b$, for some $C \in \R$.
We now investigate cases according to the sign of $b$. 
Let $b = 0$.
We have: 
{{begin-eqn}}
{{eqn | l = \map I a
      | r = \pi \map \ln {a + \sqrt {a^2} } + C
}}
{{eqn | r = \pi \map \ln {a + \size a} + C
      | c = {{Defof|Absolute Value|index = 2}}
}}
{{eqn | r = \pi \map \ln {2 a} + C
      | c = since $a \ge 0$
}}
{{end-eqn}}
On the other hand: 
{{begin-eqn}}
{{eqn | l = \map I a
      | r = \int_0^\pi \ln a \rd x
}}
{{eqn | r = \pi \ln a
      | c = Primitive of Constant
}}
{{end-eqn}}
So, by Sum of Logarithms:
:$\pi \ln 2 + \pi \ln a + C = \pi \ln a$
so:
:$C = -\pi \ln 2$
giving: 
{{begin-eqn}}
{{eqn | l = \int_0^\pi \map \ln {a + b \cos x} \rd x 
      | r = \pi \map \ln {a + \sqrt {a^2 - b^2} } - \pi \ln 2
}}
{{eqn | r = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}
      | c = Difference of Logarithms
}}
{{end-eqn}}
in the case $b = 0$.
{{qed|lemma}}
Let $b > 0$.
Then: 
{{begin-eqn}}
{{eqn | l = \map I b
      | r = \pi \map \ln {b + \sqrt {b^2 - b^2} } + C
}}
{{eqn | r = \pi \ln b + C
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \map I b
      | r = \int_0^\pi \map \ln {b + b \cos x} \rd x
}}
{{eqn | r = \int_0^\pi \ln b \rd x + \int_0^\pi \map \ln {1 + \cos x} \rd x
      | c = Sum of Logarithms
}}
{{eqn | r = \pi \ln b + \int_0^\pi \map \ln {2 \cos^2 \frac x 2} \rd x
      | c = Primitive of Constant, {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{eqn | r = \pi \ln b + \int_0^\pi \ln 2 \rd x + 2 \int_0^\pi \map \ln {\cos \frac x 2} \rd x
      | c = Sum of Logarithms, Logarithm of Power
}}
{{eqn | r = \pi \ln b + \pi \ln 2 + 4 \int_0^{\frac \pi 2} \map \ln {\cos u} \rd u
      | c = substituting $u = \dfrac x 2$
}}
{{eqn | r = \pi \ln b + \pi \ln 2 - 2 \pi \ln 2
      | c = Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\cos x}$
}}
{{eqn | r = \pi \ln b - \pi \ln 2
}}
{{end-eqn}}
giving: 
:$C = -\pi \ln 2$
So we have: 
:$\ds \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
in the case $b > 0$ too.
{{qed|lemma}}
Let $b < 0$.
We have: 
{{begin-eqn}}
{{eqn | l = \map I {\size b}
      | r = \pi \map \ln {\size b + \sqrt {\size b^2 - b^2} } + C
}}
{{eqn | r = \pi \map \ln {\size b} + C
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \map I {\size b}
      | r = \int_0^\pi \map \ln {\size b + b \cos x} \rd x
}}
{{eqn | r = \int_0^\pi \map \ln {\size b} \rd x + \int_0^\pi \map \ln {1 + \frac b {\size b} \cos x} \rd x
      | c = Sum of Logarithms
}}
{{eqn | r = \pi \ln {\size b} + \int_0^\pi \map \ln {1 - \cos x} \rd x
      | c = since $\dfrac b {\size b} = -1$ for $b < 0$ and using Primitive of Constant
}}
{{eqn | r = \pi \ln {\size b} + \int_0^\pi \map \ln {2 \sin^2 \frac x 2} \rd x
      | c = {{Corollary|Double Angle Formula for Cosine|2}}
}}
{{eqn | r = \pi \ln \size b + \int_0^\pi \ln 2 \rd x + \int_0^\pi \map \ln {\sin^2 \frac x 2} \rd x
      | c = Sum of Logarithms
}}
{{eqn | r = \pi \ln \size b + \pi \ln 2 + 2 \int_0^\pi \map \ln {\sin \frac x 2} \rd x
      | c = Primitive of Constant, Logarithm of Power
}}
{{eqn | r = \pi \ln \size b + \pi \ln 2 + 4 \int_0^{\pi/2} \map \ln {\sin u} \rd u
      | c = substituting $u = \dfrac x 2$
}}
{{eqn | r = \pi \ln \size b + \pi \ln 2 - 2 \pi \ln 2
      | c = Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$
}}
{{eqn | r = \pi \ln \size b - \pi \ln 2
}}
{{end-eqn}}
We have in this case: 
:$C = -\pi \ln 2$
So we have: 
:$\ds \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
in the case $b < 0$ too.
{{qed|lemma}}
The result follows.
{{qed}}