Integral Resulting in Arcsecant

Theorem

 * $\displaystyle \int \frac 1 {x \sqrt{x^2 - a^2} }\ \mathrm dx = \begin{cases}

\dfrac 1 {\left\vert{a}\right\vert} \operatorname {arcsec} \dfrac x {\left\vert{a}\right\vert} + C & : x > \left\vert{a}\right\vert \\ -\dfrac 1 {\left\vert{a}\right\vert} \operatorname {arcsec} \dfrac x {\left\vert{a}\right\vert} + C & : x < -\left\vert{a}\right\vert \end{cases}$

where $a$ is a constant.

Proof
Substitute:


 * $\sec \theta = \dfrac x {\left\vert{a}\right\vert} \iff \left\vert{a}\right\vert \sec \theta = x$

for $\theta \in \left({0 \,.\,.\, \dfrac \pi 2}\right) \cup \left({\dfrac \pi 2 \,.\,.\, \pi}\right)$.

This substitution is valid for all $\dfrac x {|a|} \in \R \setminus \left({-1 \,.\,.\, 1}\right)$.

By hypothesis:

so this substitution will not change the domain of the integrand.

Thus:

and so:

By Shape of Tangent Function and the stipulated definition of $\theta$:


 * $(A): \quad \dfrac x {\left\vert{a}\right\vert} > 1 \iff \theta \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$

and


 * $(B): \quad \dfrac x {\left\vert{a}\right\vert} < -1 \iff \theta \in \left({\dfrac \pi 2 \,.\,.\, \pi}\right)$

If $(A)$:

If $(B)$:

Also see

 * Derivative of Arcsecant Function