Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x/Mistake

Source Work

 * Chapter $14$: Indefinite Integrals
 * Integrals involving $\cosh a x$: $14.584$

This mistake can be seen in the edition as published by Schaum: ISBN 0-07-060224-7 (unknown printing).

Mistake

 * $\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \begin {cases}

\dfrac 1 {2 a p \sqrt {p^2 + q^2} } \map \ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } } \\ \dfrac 1 {a p \sqrt {p^2 + q^2} } \arctan \dfrac {p \tanh a x} {\sqrt {p^2 + q^2} } \end {cases}$

Correction
As demonstrated in Primitive of $\dfrac {\d x} {p^2 + q^2 \cosh^2 a x}$, the second expression is incorrect.

It should just be:


 * $\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \map \ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }$

The second expression is not completely wrong, however.

The result can also be presented as:


 * $\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \dfrac 1 {a p \sqrt {p^2 + q^2} } \tanh^{-1} \dfrac {p \tanh a x} {\sqrt {p^2 + q^2} }$

but it is a genuine equivalence which is valid over the whole range.

Hence presenting it as two separate cases, each expected to valid over a subdomain of $\R$ (in this case unspecified), is misleading.