Primitive of Power of x by Arcsecant of x over a
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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 1
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 |
$\blacksquare$
Proof 2
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:
\(\ds u\) | \(=\) | \(\ds \arcsec \frac x a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \begin {cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ \dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\ \end {cases}\) | Derivative of $\arcsec \dfrac x a$ |
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 $\arcsec \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
\(\ds \int x^m \arcsec \frac x a \rd x\) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arcsec \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arcsec \frac x a - \frac a {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - a^2} }\) | Primitive of Constant Multiple of Function |
Similarly, let $\arcsec \dfrac x a$ be in the interval $\openint {\dfrac \pi 2} \pi$.
Then:
\(\ds \int x^m \arcsec \frac x a \rd x\) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arcsec \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arcsec \frac x a + \frac a {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - a^2} }\) | Primitive of Constant Multiple of Function |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.507$