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 $u = e^{a x}$ |
$\blacksquare$
Also see
- Primitive of $\dfrac 1 {\cosh a x + 1}$ for the case where $p = q$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\cosh a x$: $14.581$