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

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

Let:

\(\ds u\) \(=\) \(\ds e^{a x}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d u} {\d x}\) \(=\) \(\ds e^{a x} = u\)
\(\ds \leadsto \ \ \) \(\ds \d x\) \(=\) \(\ds \dfrac {\d u} u\)


Hence:

\(\ds \int \frac {\d x} {p + q \sinh a x}\) \(=\) \(\ds \int \frac {\d x} {p + q \paren {\dfrac {e^{a x} - e^{-a x} } 2} }\) Definition of Real Hyperbolic Sine
\(\ds \) \(=\) \(\ds \int \frac {2 \rd x} {2 p + q \paren {e^{a x} - e^{-a x} } }\)
\(\ds \) \(=\) \(\ds \int \frac {2 \rd u} {u \paren {2 p + q u - \frac q u} }\) Integration by Substitution: $u = e^{a x}$
\(\ds \) \(=\) \(\ds \int \frac {2 \rd u} {q u^2 + 2 p u - q}\) simplifying


The discriminant of $q u^2 + 2 p u - q$ is given by:

\(\ds \map {\operatorname {Disc} } {q u^2 + 2 p u - q}\) \(=\) \(\ds \paren {2 p}^2 + 4 q^2\)
\(\ds \) \(=\) \(\ds 4 \paren {p^2 + q^2}\)

Hence the sign of $\map {\operatorname {Disc} } {q u^2 + 2 p u - q}$ is always positive.


So:

\(\ds \int \frac {2 \rd u} {q u^2 + 2 p u - q}\) \(=\) \(\ds \dfrac 2 {\sqrt {4 p^2 + 4 q^2} } \ln \size {\dfrac {2 q u + 2 p - \sqrt {4 p^2 + 4 q^2} } {2 q u + 2 p + \sqrt {4 p^2 + 4 q^2} } } + C\) Primitive of $\dfrac 1 {a x^2 + b x + c}$
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {p^2 + q^2} } \ln \size {\dfrac {q u + p - \sqrt {p^2 + q^2} } {q u + p + \sqrt {p^2 + q^2} } } + C\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {p^2 + q^2} } \ln \size {\dfrac {q e^{a x} + p - \sqrt {p^2 + q^2} } {q e^{a x} + p + \sqrt {p^2 + q^2} } } + C\) substituting $u = e^{a x}$

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_p_plus_q_by_Hyperbolic_Sine_of_a_x&oldid=504113"