Primitive of Power of x by Arccosecant of x
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Theorem
- $\ds \int x^m \arccsc x \rd x = \begin {cases}
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x + \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \\ \ds \dfrac {x^{m + 1} } {m + 1} \arccsc x - \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \\ \end {cases}$
Proof
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:
| \(\ds u\) | \(=\) | \(\ds \arccsc x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \begin {cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \\
\dfrac 1 {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \\ \end{cases}\) |
Derivative of $\arccsc x$ |
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 $\arccsc x$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
| \(\ds \int x^m \arccsc x \rd x\) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arccsc x - \int \frac {x^{m + 1} } {m + 1} \paren {\frac {-1} {x \sqrt {x^2 - 1} } } \rd x\) | Integration by Parts | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arccsc x + \frac 1 {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - 1} }\) | Primitive of Constant Multiple of Function |
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.
Then:
| \(\ds \int x^m \arccsc x \rd x\) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arccsc x - \int \frac {x^{m + 1} } {m + 1} \paren {\frac 1 {x \sqrt {x^2 - 1} } } \rd x\) | Integration by Parts | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arccsc x - \frac 1 {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - 1} }\) | Primitive of Constant Multiple of Function |
$\blacksquare$