Primitive of Reciprocal of p squared minus Square of q by Hyperbolic Sine of a x

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Theorem

$\ds \int \frac {\d x} {p^2 - q^2 \sinh^2 a x} = \frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\frac {p + \sqrt {p^2 + q^2} \tanh a x} {p - \sqrt {p^2 + q^2} \tanh a x} } + C$


Proof

\(\ds \int \frac {\d x} {p^2 - q^2 \sinh^2 a x}\) \(=\) \(\ds \int \frac {\sech^2 a x \rd x} {p^2 \sech^2 a x - q^2 \tanh^2 a x}\) multiplying numerator and denominator by $\sech^2 a x$
\(\ds \) \(=\) \(\ds \int \frac {\sech^2 a x \rd x} {p^2 \paren {1 - \tanh^2 a x} - q^2 \tanh^2 a x}\) Sum of Squares of Hyperbolic Secant and Tangent
\(\ds \) \(=\) \(\ds \int \frac {\sech^2 a x \rd x} {p^2 - \paren {p^2 + q^2} \tanh^2 a x}\)


Then:

\(\ds u\) \(=\) \(\ds \tanh a x\)
\(\ds \leadsto \ \ \) \(\ds \d u\) \(=\) \(\ds a \sech^2 a x \rd x\) Derivative of Hyperbolic Tangent Function
\(\ds \leadsto \ \ \) \(\ds \int \frac {\sech^2 a x \rd x} {p^2 - \paren {p^2 + q^2} \tanh^2 a x}\) \(=\) \(\ds \int \frac {\d u} {a p^2 - a \paren {p^2 + q^2} u^2}\) Integration by Substitution: $u = \tanh a x$
\(\ds \) \(=\) \(\ds \frac 1 {a \paren {p^2 + q^2} } \int \frac {\d u} {\paren {\frac {p^2} {p^2 + q^2} } - u^2}\)
\(\ds \) \(=\) \(\ds \frac {-1} {a \paren {p^2 + q^2} } \int \frac {\d u} {u^2 - \paren {\frac p {\sqrt {p^2 + q^2} } }^2}\)
\(\ds \) \(=\) \(\ds \frac {-1} {a \paren {p^2 + q^2} } \frac {\sqrt {p^2 + q^2} } {2 p} \ln \size {\frac {u - \frac p {\sqrt {p^2 + q^2} } } {u + \frac p {\sqrt {p^2 + q^2} } } } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$
\(\ds \) \(=\) \(\ds \frac {-1} {2 a p \sqrt {p^2 + q^2} } \ln \size {\frac {\sqrt {p^2 + q^2} u - p} {\sqrt {p^2 + q^2} u + p} } + C\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\frac {\sqrt {p^2 + q^2} u + p} {\sqrt {p^2 + q^2} u - p} } + C\) Logarithm of Reciprocal
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\frac {\sqrt {p^2 + q^2} \tanh a x + p} {\sqrt {p^2 + q^2} \tanh a x - p} } + C\) substituting $u = \tanh a x$
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\frac {p + \sqrt {p^2 + q^2} \tanh a x} {p - \sqrt {p^2 + q^2} \tanh a x} } + C\) as $\size {a - b} - \size {b - a}$

$\blacksquare$


Also see


Sources

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