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

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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:

\(\ds \map {I'} \alpha\) \(=\) \(\ds \frac \d {\d \alpha} \int_0^{\pi/2} \sec x \map \ln {1 + \alpha \cos x} \rd x\)
\(\ds \) \(=\) \(\ds \int_0^{\pi/2} \frac \partial {\partial \alpha} \paren {\sec x \map \ln {1 + \alpha \cos x} } \rd x\) Definite Integral of Partial Derivative
\(\ds \) \(=\) \(\ds \int_0^{\pi/2} \frac {\cos x \sec x} {1 + \alpha \cos x} \rd x\) Chain Rule for Derivatives, Derivative of Natural Logarithm
\(\ds \) \(=\) \(\ds \int_0^{\pi/2} \frac 1 {1 + \alpha \cos x} \rd x\)
\(\ds \) \(=\) \(\ds \frac {\arccos \alpha} {\sqrt {1 - \alpha^2} }\) Definite Integral from $0$ to $\dfrac \pi 2$ of $\dfrac 1 {a + b \cos x}$

Then:

\(\ds \map I b - \map I a\) \(=\) \(\ds \int_a^b \map {I'} \alpha \rd \alpha\) Fundamental Theorem of Calculus: Second Part
\(\ds \) \(=\) \(\ds \int_a^b \frac {\arccos \alpha} {\sqrt {1 - \alpha^2} } \rd \alpha\)

From Derivative of Arccosine Function, we have:

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

so:

\(\ds \int_a^b \frac {\arccos \alpha} {\sqrt {1 - \alpha^2} } \rd \alpha\) \(=\) \(\ds \int_{\arccos a}^{\arccos b} x \frac {-\sqrt {1 - \alpha^2} } {\sqrt {1 - \alpha^2} } \rd x\) substituting $x = \arccos \alpha$
\(\ds \) \(=\) \(\ds -\int_{\arccos a}^{\arccos b} x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {-\frac {x^2} 2} {\arccos a} {\arccos b}\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2}\)

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.

$\blacksquare$


Sources