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

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Theorem

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


Proof

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

Hence the result.

$\blacksquare$


Also see


Sources

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