Primitive of Power of x by Arccosecant of x over a/Proof 1

Theorem

$\ds \int x^m \arccsc \frac x a \rd x = \begin {cases} \ds \dfrac {x^{m + 1} } {m + 1} \arccsc \dfrac x a + \dfrac a {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\ \\ \ds \dfrac {x^{m + 1} } {m + 1} \arccsc \dfrac x a - \dfrac a {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \end {cases}$

Proof

Recall:

$\text {(1)}: \quad$

$\ds \int x^m \arccsc x \rd x$

$=$

$\ds \begin {cases} \ds \dfrac {x^{m + 1} } {m + 1} \arccsc x + \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \\ \\ \ds \dfrac {x^{m + 1} } {m + 1} \arccsc x - \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \\ \end {cases} $

Primitive of $x^m \arccsc x$

Then:

$\ds \int x^m \arccsc \frac x a \rd x$

$=$

$\ds \int a^m \paren {\dfrac x a}^m \arccsc \frac x a \rd x$

manipulating into appropriate form

$\ds $

$=$

$\ds a^m \int \paren {\dfrac x a}^m \arccsc \frac x a \rd x$

Primitive of Constant Multiple of Function

$\ds $

$=$

$\ds a^m \paren {\dfrac 1 {1 / a} \paren {\begin {cases} \dfrac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arccsc \dfrac x a + \dfrac 1 {m + 1} \ds \int \paren {\dfrac x a}^m \frac {\d x} {\sqrt {\paren {\dfrac x a}^2 - 1} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\ \\ \dfrac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arccsc \dfrac x a - \dfrac 1 {m + 1} \ds \int \paren {\dfrac x a}^m \frac {\d x} {\sqrt {\paren {\dfrac x a}^2 - 1} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \end {cases} } } $

Primitive of Function of Constant Multiple, from $(1)$

$\ds $

$=$

$\ds \begin {cases} \dfrac {x^{m + 1} } {m + 1} \arccsc \dfrac x a + \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\ \\ \dfrac {x^{m + 1} } {m + 1} \arccsc \dfrac x a - \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \end {cases} $

simplifying

$\blacksquare$