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

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int \frac {\d x} {\paren {p + q \cosh a x}^2} = \frac {q \sinh a x} {a \paren {q^2 - p^2} \paren {p + q \cosh a x} } - \frac p {q^2 - p^2} \int \frac {\rd x} {p + q \cosh a x} + C$


Proof

\(\ds \map {\dfrac \d {\d x} } {\dfrac {\sinh a x} {p + q \cosh a x} }\) \(=\) \(\ds \dfrac {\paren {p + q \cosh a x} \map {\frac \d {\d x} } {\sinh a x} - \sinh a x \map {\frac \d {\d x} } {p + q \sinh a x} } {\paren {p + q \cosh a x}^2}\) Quotient Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac {\paren {p + q \sinh a x} \paren {a \cosh a x} - \sinh a x \paren {a q \sinh a x} } {\paren {p + q \cosh a x}^2}\) Derivative of Hyperbolic Cosine Function, Derivative of Hyperbolic Sine Function
\(\ds \) \(=\) \(\ds a \dfrac {p \cosh a x + q \paren {\cosh^2 a x - \sinh^2 a x} } {\paren {p + q \cosh a x}^2}\) simplification
\(\ds \) \(=\) \(\ds a \dfrac {p \cosh a x + q} {\paren {p + q \cosh a x}^2}\) Difference of Squares of Hyperbolic Cosine and Sine
\(\ds \) \(=\) \(\ds a \dfrac {p q \cosh a x + q^2} {q \paren {p + q \cosh a x}^2}\)
\(\ds \) \(=\) \(\ds a \dfrac {p q \cosh a x + q^2 + p^2 - p^2} {q \paren {p + q \cosh a x}^2}\)
\(\ds \) \(=\) \(\ds a \dfrac {q^2 - p^2} {q \paren {p + q \cosh a x}^2} + a \dfrac {p q \cosh a x + p^2} {q \paren {p + q \cosh a x}^2}\)
\(\ds \) \(=\) \(\ds a \dfrac {q^2 - p^2} {q \paren {p + q \cosh a x}^2} + a \dfrac {p \paren {p + q \cosh a x} } {q \paren {p + q \cosh a x}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {a \paren {q^2 - p^2} } {q \paren {p + q \cosh a x}^2} + \dfrac {a p} q \dfrac 1 {p + q \cosh a x}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac q {a \paren {q^2 - p^2} } \map {\dfrac \d {\d x} } {\dfrac {\sinh a x} {p + q \cosh a x} }\) \(=\) \(\ds \dfrac 1 {\paren {p + q \cosh a x}^2} + \dfrac p {\paren {q^2 - p^2} } \dfrac 1 {p + q \cos a x}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac q {a \paren {q^2 - p^2} } {\dfrac {\sinh a x} {p + q \cosh a x} }\) \(=\) \(\ds \int \dfrac {\d x} {\paren {p + q \cosh a x}^2} + \dfrac p {\paren {q^2 - p^2} } \int \dfrac {\d x} {p + q \cosh a x}\)

Hence the result.

$\blacksquare$


Also see


Sources