Primitive of Power of x by Arccosecant of x over a

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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 1

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

$\blacksquare$


Proof 2

With a view to expressing the primitive in the form:

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

let:

\(\ds u\) \(=\) \(\ds \arccsc \frac x a\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \begin {cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\

\dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\ \end{cases}\)

Derivative of $\arccsc \dfrac x a$


and let:

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


First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.

Then:

\(\ds \int x^m \arccsc \frac x a \rd x\) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \arccsc \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \arccsc \frac x a + \frac a {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - a^2} }\) Primitive of Constant Multiple of Function


Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.

Then:

\(\ds \int x^m \arccsc \frac x a \rd x\) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \arccsc \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \arccsc \frac x a - \frac a {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - a^2} }\) Primitive of Constant Multiple of Function

$\blacksquare$


Also see


Sources

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