Integral Resulting in Arcsecant

Theorem

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

\dfrac 1 {|a|} \text {arcsec} \dfrac x {|a|} + C& : x > |a| \\ -\dfrac {1} {|a|} \text {arcsec} \dfrac x {|a|} + C& : x < -|a| \end{cases}$

where $a$ is a constant.

Proof
Substitute:


 * $\sec \theta = \dfrac x {|a|} \iff |a|\sec \theta = x$

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:


 * $\left(x > |a|\right) \lor \left(x < -|a|\right)$

$\iff \left(\dfrac {x}{|a|} > 1\right) \lor \left(\dfrac {x}{|a|} < -1\right)$

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

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


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

and


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

If $(A)$:

If $(B)$:

Also see

 * Derivative of Arcsecant Function