Integral Resulting in Arcsecant

Theorem

 * $\ds \int \frac 1 {x \sqrt {x^2 - a^2} } \rd x = \begin{cases}

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

where $a$ is a constant.

Proof
Substitute:


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

for $\theta \in \openint 0 {\dfrac \pi 2} \cup \openint {\dfrac \pi 2} \pi$.

This substitution is valid for all $\dfrac x {\size a} \in \R \setminus \openint {-1} 1$.

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 $\theta$:


 * $(A): \quad \dfrac x {\size a} > 1 \iff \theta \in \openint 0 {\dfrac \pi 2}$

and


 * $(B): \quad \dfrac x {\size a} < -1 \iff \theta \in \openint {\dfrac \pi 2} \pi$

If $(A)$:

If $(B)$:

Also see

 * Derivative of Arcsecant Function