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