Primitive of Power of x by Inverse Hyperbolic Cosecant of x over a

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int x^m \arcsch \frac x a \rd x = \begin{cases}

\ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a + \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x > 0 \\ \ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a - \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x < 0 \\ \end{cases}$


Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \arcsch \frac x a\)
\(\ds \leadsto \ \ \) \(\ds \frac {\rd u} {\rd x}\) \(=\) \(\ds \frac {-a} {\size x \sqrt{a^2 + x^2} }\) Derivative of $\arcsch \dfrac x a$


and let:

\(\ds \frac {\rd v} {\rd x}\) \(=\) \(\ds x^m\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {x^{m + 1} } {m + 1}\) Primitive of Power


Then:

\(\ds \int \frac {\arcsch \dfrac x a \rd x} {x^2}\) \(=\) \(\ds \paren {\arcsch \frac x a} \paren {\frac {x^{m + 1} } {m + 1} } - \int \paren {\frac {x^{m + 1} } {m + 1} } \paren {\frac {-a} {\size x \sqrt{a^2 + x^2} } } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \begin {cases}

\ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a + \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x > 0 \\ \ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a - \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x < 0 \\ \end {cases}\)

Definition of Absolute Value

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Power_of_x_by_Inverse_Hyperbolic_Cosecant_of_x_over_a&oldid=505096"