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

From ProofWiki
Jump to navigation Jump to search

Theorem

$\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} } \ln \size {\dfrac {p \tanh a x + \sqrt {p^2 - q^2} } {p \tanh a x - \sqrt {p^2 - q^2} } } + C & : p^2 > q^2 \\ \dfrac 1 {a p \sqrt {q^2 - p^2} } \arctan \dfrac {p \tanh a x} {\sqrt {q^2 - p^2} } + C & : p^2 < q^2 \\ \end {cases}$


Proof

\(\ds \int \frac {\d x} {p^2 - q^2 \cosh^2 a x}\) \(=\) \(\ds \int \frac {\csch^2 a x \rd x} {p^2 \csch^2 a x - q^2 \coth^2 a x}\) multiplying numerator and denominator by $\csch^2 a x$
\(\ds \) \(=\) \(\ds \int \frac {\csch^2 a x \rd x} {p^2 \paren {\coth^2 a x - 1} - q^2 \coth^2 a x}\) Difference of Squares of Hyperbolic Cotangent and Cosecant
\(\ds \) \(=\) \(\ds \int \frac {\csch^2 a x \rd x} {\paren {p^2 - q^2} \coth^2 a x - p^2}\) simplifying

Let:

\(\ds u\) \(=\) \(\ds \coth a x\)
\(\ds \leadsto \ \ \) \(\ds \d u\) \(=\) \(\ds -a \csch^2 a x \rd x\) Derivative of Hyperbolic Cotangent Function
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {p^2 - q^2 \cosh^2 a x}\) \(=\) \(\ds \frac 1 a \int \frac {-\d u} {\paren {p^2 - q^2} u^2 - p^2}\) Integration by Substitution: substituting $u = \tanh a x$
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac 1 {a \paren {p^2 - q^2} } \int \frac {\rd u} {\frac {p^2} {p^2 - q^2} - u^2}\) rearranging into a standard form


There are two cases to address.


First, suppose $p^2 > q^2$.

Then we have that $p^2 - q^2 > 0$, and so:

\(\ds \int \frac {\d x} {p^2 - q^2 \cosh^2 a x}\) \(=\) \(\ds \frac 1 {a \paren {p^2 - q^2} } \int \frac {\rd u} {\paren {\frac p {\sqrt {p^2 - q^2} } }^2 - u^2}\) from $(1)$
\(\ds \) \(=\) \(\ds \frac 1 {a \paren {p^2 - q^2} } \frac {\sqrt {p^2 - q^2} } {2 p} \ln \size {\frac {\frac p {\sqrt {p^2 - q^2} } + u} {\frac p {\sqrt {p^2 - q^2} } - u} } + C\) Primitive of $\dfrac 1 {a^2 - x^2}$
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {p^2 - q^2} } \ln \size {\frac {\frac p {\sqrt {p^2 - q^2} } + \coth a x} {\frac p {\sqrt {p^2 - q^2} } - \coth a x} } + C\) substituting $u = \coth a x$
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {p^2 - q^2} } \ln \size {\frac {\frac p {\sqrt {p^2 - q^2} } + \frac 1 {\tanh a x} } {\frac p {\sqrt {p^2 - q^2} } - \frac 1 {\tanh a x} } } + C\) Definition of Real Hyperbolic Cotangent
\(\ds \) \(=\) \(\ds \frac 1 {2 a p \sqrt {p^2 - q^2} } \ln \size {\frac {p \tanh a x + \sqrt {p^2 - q^2} } {p \tanh a x - \sqrt {p^2 - q^2} } } + C\) multiplying numerator and denominator of argument of $\ln$ by $\sqrt {p^2 - q^2} \tanh a x$


Now suppose $p^2 < q^2$.

Then we have that $p^2 - q^2 < 0$, and so:

\(\ds \int \frac {\d x} {p^2 - q^2 \cosh^2 a x}\) \(=\) \(\ds \frac 1 {a \paren {p^2 - q^2} } \int \frac {\rd u} {\frac {p^2} {p^2 - q^2} - u^2}\) from $(1)$
\(\ds \) \(=\) \(\ds \frac 1 {a \paren {p^2 - q^2} } \int \frac {\rd u} {\frac {-p^2} {q^2 - p^2} - u^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {a \paren {q^2 - p^2} } \int \frac {\rd u} {u^2 + \paren {\frac p {\sqrt {q^2 - p^2} } }^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {a \paren {q^2 - p^2} } \frac {\sqrt {q^2 - p^2} } p \arctan \frac u {\frac p {\sqrt {q^2 - p^2} } } + C\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {q^2 - p^2} } \arctan \frac {\sqrt {q^2 - p^2} u} p + C\)
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {q^2 - p^2} } \arctan \frac {\sqrt {q^2 - p^2} } p \coth a x + C\) substituting $u = \coth a x$
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {q^2 - p^2} } \arctan \frac {\sqrt {q^2 - p^2} {p \tanh a x} } + C\) Definition of Real Hyperbolic Cotangent
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {q^2 - p^2} } \arccot \frac {p \tanh a x} {\sqrt {q^2 - p^2} } + C\) Arctangent of Reciprocal equals Arccotangent
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {q^2 - p^2} } \map \arctan {\frac {p \tanh a x} {\sqrt {q^2 - p^2} } - \dfrac \pi 2} + C\) Sum of Arctangent and Arccotangent
\(\ds \) \(=\) \(\ds \frac 1 {a p \sqrt {q^2 - p^2} } \arctan \frac {p \tanh a x} {\sqrt {q^2 - p^2} } + C\) subsuming $\dfrac \pi 2$ into constant

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_p_squared_minus_Square_of_q_by_Hyperbolic_Cosine_of_a_x&oldid=504135"