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