Primitive of Reciprocal of p squared minus square of q by Cosine of a x

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Theorem

$\ds \int \frac {\rd x} {p^2 - q^2 \cos^2 a x} = \begin {cases}

\dfrac 1 {a p \sqrt {p^2 - q^2} } \arctan \dfrac {p \tan a x} {\sqrt {p^2 - q^2} } & : p^2 > q^2 \\ \dfrac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\dfrac {p \tan a x - \sqrt {q^2 - p^2} } {p \tan a x + \sqrt {q^2 - p^2} } } & : p^2 < q^2 \end {cases}$


Proof

\(\ds \int \frac {\d x} {p^2 - q^2 \cos^2 a x}\) \(=\) \(\ds \int \frac {\csc^2 a x \rd x} {p^2 \csc^2 a x - q^2 \cot^2 a x}\) multiplying numerator and denominator by $\csc^2 a x$
\(\ds \) \(=\) \(\ds \int \frac {\csc^2 a x \rd x} {p^2 \paren {1 + \cot^2 a x} - q^2 \cot^2 a x}\) Difference of Squares of Cosecant and Cotangent
\(\ds \) \(=\) \(\ds \int \frac {\csc^2 a x \rd x} {p^2 + \paren {p^2 - q^2} \cot^2 a x}\) simplifying

Let:

\(\ds t\) \(=\) \(\ds \cot a x\)
\(\ds \leadsto \ \ \) \(\ds \d t\) \(=\) \(\ds -a \csc^2 a x \rd x\) Derivative of Cotangent Function
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {p^2 - q^2 \cos^2 a x}\) \(=\) \(\ds \frac 1 a \int \frac {-\d t} {p^2 + \paren {p^2 - q^2} t^2}\) Integration by Substitution: substituting $t = \cot a x$
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac {-1} {a \paren {p^2 - q^2} } \int \frac {\rd t} {t^2 + \tfrac {p^2} {p^2 - q^2} }\)


There are two cases to address.


First, suppose $p^2 > q^2$.

Then we have that $p^2 - q^2 > 0$, and so:

\(\ds \int \frac {\d x} {p^2 - q^2 \cos^2 a x}\) \(=\) \(\ds \frac {-1} {a \paren {p^2 - q^2} } \int \frac {\rd t} {t^2 + \paren {\tfrac p {\sqrt {p^2 - q^2} } }^2}\) from $(1)$
\(\ds \) \(=\) \(\ds \frac {-1} {a \paren {p^2 - q^2} } \frac {\sqrt {p^2 - q^2} } p \map \arctan {\frac {\sqrt {p^2 - q^2} } p t} + C\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac {-1} {a p \sqrt {p^2 - q^2} } \map \arctan {\frac {\sqrt {p^2 - q^2} } p t} + C\)
\(\ds \) \(=\) \(\ds \frac {-1} {a p \sqrt {p^2 - q^2} } \map \arctan {\frac {\sqrt {p^2 - q^2} } p \cot a x} + C\) substituting $t = \cot a x$
\(\ds \) \(=\) \(\ds \frac {-1} {a p \sqrt {p^2 - q^2} } \map \arctan {\frac {\sqrt {p^2 - q^2} } {p \tan a x} } + C\) Definition of Real Cotangent Function
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {p^2 - q^2} } \map \arccot {\frac {p \tan a x} {\sqrt {p^2 - q^2} } } + C\) Arctangent of Reciprocal equals Arccotangent
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {p^2 - q^2} } \paren {\map \arctan {\frac {p \tan a x} {\sqrt {p^2 - q^2} } } - \dfrac \pi 2} + C\) Sum of Arctangent and Arccotangent
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {p^2 - q^2} } \map \arctan {\frac {p \tan a x} {\sqrt {p^2 - q^2} } } + C\) subsuming $\dfrac \pi 2$ into constant


Now suppose $p^2 < q^2$.

Then we have that $p^2 - q^2 < 0$, and so:


\(\ds \int \frac {\d x} {p^2 - q^2 \cos^2 a x}\) \(=\) \(\ds \frac {-1} {a \paren {p^2 - q^2} } \int \frac {\rd t} {t^2 + \tfrac {p^2} {p^2 - q^2} }\) from $(1)$
\(\ds \) \(=\) \(\ds \frac {-1} {a \paren {p^2 - q^2} } \int \frac {\rd t} {t^2 - \tfrac {p^2} {q^2 - p^2} }\)
\(\ds \) \(=\) \(\ds \frac {-1} {a \paren {p^2 - q^2} } \int \frac {\rd t} {t^2 - \paren {\tfrac p {\sqrt {q^2 - p^2} } }^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {a \paren {q^2 - p^2} } \frac {\sqrt {q^2 - p^2} } {2 p} \ln \size {\frac {t - \tfrac p {\sqrt {q^2 - p^2} } } {t + \tfrac p {\sqrt {q^2 - p^2} } } } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\frac {\cot a x - \tfrac p {\sqrt {q^2 - p^2} } } {\cot a x + \tfrac p {\sqrt {q^2 - p^2} } } } + C\) substituting $t = \cot a x$
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\frac {\tfrac 1 {\tan a x} - \tfrac p {\sqrt {q^2 - p^2} } } {\tfrac 1 {\tan a x} + \tfrac p {\sqrt {q^2 - p^2} } } } + C\) Definition of Real Cotangent Function
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\frac {\sqrt {q^2 - p^2} - p \tan a x} {\sqrt {q^2 - p^2} + p \tan a x} } + C\) multiplying top and bottom of argument of $\ln$ by $\sqrt {q^2 - p^2} \tan a x$
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\frac {p \tan a x - \sqrt {q^2 - p^2} } {p \tan a x + \sqrt {q^2 - p^2} } } + C\) as $\size {a - b} = \size {b - a}$

$\blacksquare$


Also see


Sources

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