Definite Integral from 0 to Pi of Sec x by Logarithm of One plus b Cosine x over One plus a Cosine x

Theorem

 * $\ds \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2}$

where $a$ and $b$ are real numbers with $-1 < a, b < 1$.

Proof
Note that by Difference of Logarithms:


 * $\ds \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \int_0^{\pi/2} \sec x \map \ln {1 + b \cos x} \rd x - \int_0^{\pi/2} \sec \map \ln {1 + a \cos x} \rd x$

For each $\alpha \in \openint {-1} 1$, set:


 * $\ds \map I \alpha = \int_0^{\pi/2} \sec x \map \ln {1 + \alpha \cos x} \rd x$

Then:


 * $\ds \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \map I b - \map I a$

We have:

Then:

From Derivative of Arccosine Function, we have:


 * $\ds \frac \d {\d \alpha} \arccos \alpha = -\frac 1 {\sqrt {1 - \alpha^2} }$

so:

so:


 * $\ds \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2}$

as required.