Primitive of Power of x by Arcsecant of x over a/Proof 1
Theorem
- $\ds \int x^m \arcsec \frac x a \rd x = \begin {cases}
\dfrac {x^{m + 1} } {m + 1} \arcsec \dfrac x a - \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ \dfrac {x^{m + 1} } {m + 1} \arcsec \dfrac x a + \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\ \end {cases}$
Proof
Recall:
\(\text {(1)}: \quad\) | \(\ds \int x^m \arcsec x \rd x\) | \(=\) | \(\ds \begin {cases}
\dfrac {x^{m + 1} } {m + 1} \arcsec x - \dfrac 1 {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arcsec x < \dfrac \pi 2 \\ \dfrac {x^{m + 1} } {m + 1} \arcsec x + \dfrac 1 {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : \dfrac \pi 2 < \arcsec x < \pi \\ \end {cases}\) |
Primitive of $x^m \arcsec x$ |
Then:
\(\ds \int x^m \arcsec \frac x a \rd x\) | \(=\) | \(\ds \int a^m \paren {\dfrac x a}^m \arcsec \frac x a \rd x\) | manipulating into appropriate form | |||||||||||
\(\ds \) | \(=\) | \(\ds a^m \int \paren {\dfrac x a}^m \arcsec \frac x a \rd x\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds a^m \paren {\dfrac 1 {1 / a} \paren {\begin {cases}
\dfrac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arcsec \dfrac x a - \dfrac 1 {m + 1} \ds \int \paren {\dfrac x a}^m \frac {\d x} {\sqrt {\paren {\dfrac x a}^2 - 1} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ \dfrac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arcsec \dfrac x a + \dfrac 1 {m + 1} \ds \int \paren {\dfrac x a}^m \frac {\d x} {\sqrt {\paren {\dfrac x a}^2 - 1} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\ \end {cases} } }\) |
Primitive of Function of Constant Multiple, from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin {cases}
\dfrac {x^{m + 1} } {m + 1} \arcsec \dfrac x a - \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ \dfrac {x^{m + 1} } {m + 1} \arcsec \dfrac x a + \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\ \end {cases}\) |
simplifying |
$\blacksquare$