Primitive of Arcsecant of x over a over x squared

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Theorem

$\displaystyle \int \frac {\operatorname{arcsec} \frac x a} {x^2} \ \mathrm d x = \begin{cases} \displaystyle \frac {-\operatorname{arcsec} \frac x a} x + \frac {\sqrt{x^2 - a^2} } {a x} + C & : 0 < \operatorname{arcsec} \dfrac x a < \dfrac \pi 2 \\ \displaystyle \frac {-\operatorname{arcsec} \frac x a} x - \frac {\sqrt{x^2 - a^2} } {a x} + C & : \dfrac \pi 2 < \operatorname{arcsec} \dfrac x a < \pi \\ \end{cases}$


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle \operatorname{arcsec} \frac x a\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle \begin{cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \operatorname{arcsec} \dfrac x a < \dfrac \pi 2 \\ \dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \operatorname{arcsec} \dfrac x a < \pi \\ \end{cases}\) Derivative of $\operatorname{arcsec} \dfrac x a$


and let:

\(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle \frac 1 {x^2}\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {-1} x\) Primitive of Power


First let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({0 \,.\,.\,\dfrac \pi 2}\right)$.

Then:

\(\displaystyle \int \frac {\operatorname{arcsec} \frac x a} {x^2}\) \(=\) \(\displaystyle \frac {-1} x \operatorname{arcsec} \frac x a - \int \frac {-1} x \left({\frac a {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\operatorname{arcsec} \frac x a} x + a \int \frac {\mathrm d x} {x \sqrt {x^2 - a^2} } + C\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\operatorname{arcsec} \frac x a} x + a \left({\frac {\sqrt {x^2 - a^2} } {a^2 x} }\right) + C\) Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\operatorname{arcsec} \frac x a} x + \frac {\sqrt{x^2 - a^2} } {a x} + C\) simplifying


Similarly, let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({\dfrac \pi 2 \,.\,.\, \pi}\right)$.

Then:

\(\displaystyle \int \frac {\operatorname{arcsec} \frac x a} {x^2}\) \(=\) \(\displaystyle \frac {-1} x \operatorname{arcsec} \frac x a - \int \frac {-1} x \left({\frac {-a} {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\operatorname{arcsec} \frac x a} x - a \int \frac {\mathrm d x} {x \sqrt {x^2 - a^2} } + C\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\operatorname{arcsec} \frac x a} x - a \left({\frac {\sqrt {x^2 - a^2} } {a^2 x} }\right) + C\) Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\operatorname{arcsec} \frac x a} x - \frac {\sqrt{x^2 - a^2} } {a x} + C\) simplifying

$\blacksquare$


Also see


Sources