Primitive of Reciprocal of p plus q by Hyperbolic Cosine of a x

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Theorem

For $p \ne q$:

$\ds \int \frac {\d x} {p + q \cosh a x} = \begin {cases} \dfrac 2 {a \sqrt {q^2 - p^2} } \arctan \dfrac {q e^{a x} + p} {\sqrt {q^2 - p^2} } + C & : p^2 < q^2 \\ \dfrac 1 {a \sqrt {p^2 - q^2} } \ln \size {\dfrac {q e^{a x} + p - \sqrt {p^2 - q^2} } {q e^{a x} + p + \sqrt {p^2 - q^2} } } + C & : p^2 > q^2 \end {cases}$


Proof

Let:

\(\ds u\) \(=\) \(\ds e^{a x}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d u} {\d x}\) \(=\) \(\ds e^{a x} = u\)
\(\ds \leadsto \ \ \) \(\ds \d x\) \(=\) \(\ds \dfrac {\d u} u\)


Hence:

\(\ds \int \frac {\d x} {p + q \cosh a x}\) \(=\) \(\ds \int \frac {\d x} {p + q \paren {\dfrac {e^{a x} + e^{-a x} } 2} }\) Definition of Real Hyperbolic Cosine
\(\ds \) \(=\) \(\ds \int \frac {2 \rd x} {2 p + q \paren {e^{a x} + e^{-a x} } }\)
\(\ds \) \(=\) \(\ds \int \frac {2 \rd u} {u \paren {2 p + q u + \frac q u} }\) Integration by Substitution: $u = e^{a x}$
\(\ds \) \(=\) \(\ds \int \frac {2 \rd u} {q u^2 + 2 p u + q}\) simplifying


The discriminant of $q u^2 + 2 p u + q$ is given by:

\(\ds \map {\operatorname {Disc} } {q u^2 + 2 p u + q}\) \(=\) \(\ds \paren {2 p}^2 - 4 q^2\)
\(\ds \) \(=\) \(\ds 4 \paren {p^2 - q^2}\)

Hence the result depends on the sign of $p^2 - q^2$.


Let $p^2 < q^2$.

Then:

\(\ds \int \frac {2 \rd u} {q u^2 + 2 p u + q}\) \(=\) \(\ds \dfrac 2 {\sqrt {4 q^2 - 4 p^2} } \map \arctan {\dfrac {2 q u + 2 p} {\sqrt {4 q^2 - 4 p ^2} } } + C\) Primitive of $\dfrac 1 {a x^2 + b x + c}$
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {q^2 - p^2} } \map \arctan {\dfrac {q u + p} {\sqrt {q^2 - p ^2} } } + C\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {q^2 - p^2} } \map \arctan {\dfrac {q e^{a x} + p} {\sqrt {q^2 - p^2} } } + C\) substituting $u = e^{a x}$


Let $p^2 > q^2$.

Then:

\(\ds \int \frac {2 \rd u} {q u^2 + 2 p u + q}\) \(=\) \(\ds \dfrac 2 {\sqrt {4 p^2 - 4 q^2} } \ln \size {\dfrac {2 q u + 2 p - \sqrt {4 p^2 - 4 q^2} } {2 q u + 2 p + \sqrt {4 p^2 - 4 q^2} } } + C\) Primitive of $\dfrac 1 {a x^2 + b x + c}$
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {p^2 - q^2} } \ln \size {\dfrac {q u + p - \sqrt {p^2 - q^2} } {q u + p + \sqrt {p^2 - q^2} } } + C\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {p^2 - q^2} } \ln \size {\dfrac {q e^{a x} + p - \sqrt {p^2 - q^2} } {q e^{a x} + p + \sqrt {p^2 - q^2} } } + C\) substituting $u = e^{a x}$

$\blacksquare$


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