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

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Theorem

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

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


Proof

Since $a > b > 0$, we have $a^2 > b^2$.

So:

\(\ds \int_0^{\pi/2} \frac 1 {a + b \cos x} \rd x\) \(=\) \(\ds \intlimits {\frac 2 {\sqrt {a^2 - b^2} } \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } 0 1\) Primitive of $\dfrac 1 {p + q \cos x}$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {a^2 - b^2} } \paren {2 \map \arctan {\sqrt {\frac {a - b} {a + b} } } }\)
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {a^2 - b^2} } \paren {2 \map \arctan {\sqrt {\frac {1 - \frac b a} {1 + \frac b a} } } }\)
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {a^2 - b^2} } \map \arccos {\frac b a}\) Arccosine in terms of Arctangent

$\blacksquare$


Sources

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