# 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

 $\ds \int \frac 1 {x \sqrt {x^2 - a^2} } \rd x$ $=$ $\ds \int \frac 1 {x \sqrt {a^2 \paren {\frac {x^2} {a^2} - 1} } } \rd x$ factor $a^2$ out of the radicand $\ds$ $=$ $\ds \int \frac 1 {x \sqrt {a^2} \sqrt {\paren {\frac x a}^2 - 1} } \rd x$ $\ds$ $=$ $\ds \frac 1 {\size a} \int \frac 1 {x \sqrt {\paren {\frac x a}^2 - 1} } \rd x$
$\sec \theta = \dfrac x {\size a} \iff \size a \sec \theta = x$

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

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

 $\ds \paren {x > \size a}$ $\lor$ $\ds \paren {x < - \size a}$ $\ds \leadstoandfrom \ \$ $\ds \paren {\dfrac x {\size a} > 1}$ $\lor$ $\ds \paren {\dfrac x {\size a} < -1}$

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

Thus:

 $\ds \size a \sec \theta$ $=$ $\ds x$ from above $\ds \leadsto \ \$ $\ds \size a \sec \theta \tan \theta \frac {\d \theta} {\d x}$ $=$ $\ds 1$ Differentiate with respect to $x$, Derivative of Secant Function, Chain Rule for Derivatives

and so:

 $\ds \int \frac 1 {x\sqrt {x^2 - a^2} } \rd x$ $=$ $\ds \frac 1 {\size a} \int \frac {\size a \sec \theta \tan \theta} {\size a \sec \theta \sqrt {\sec^2 \theta - 1} } \frac {\d \theta} {\d x} \rd x$ from above $\ds$ $=$ $\ds \frac 1 {\size a} \int \frac {\tan \theta} {\sqrt {\sec^2 \theta - 1} } \rd \theta$ Integration by Substitution $\ds$ $=$ $\ds \frac 1 {\size a} \int \frac {\tan \theta} {\sqrt {\tan^2 \theta} \rd \theta}$ corollary to sum of squares of sine and cosine $\ds$ $=$ $\ds \frac 1 {\size a} \int \frac {\tan \theta} {\size {\tan \theta} } \rd \theta$

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 $(A)$:

 $\ds \frac 1 {\size a} \int \frac {\tan \theta} {\size {\tan \theta} } \rd \theta$ $=$ $\ds \frac 1 {\size a} \int \rd \theta$ Definition of Absolute Value $\ds$ $=$ $\ds \frac 1 {\size a} \theta + C$ Integral of Constant $\ds$ $=$ $\ds \frac 1 {\size a} \arcsec \frac x {\size a} + C$ Definition of Arcsecant

If $(B)$:

 $\ds \frac 1 {\size a} \int \frac {\tan \theta} {\size {\tan \theta} } \rd \theta$ $=$ $\ds \frac 1 {\size a} \int -1 \rd \theta$ Definition of Absolute Value $\ds$ $=$ $\ds -\frac 1 {\size a} \theta + C$ Integral of Constant $\ds$ $=$ $\ds -\frac 1 {\size a} \arcsec \frac x {\size a} + C$ Definition of Arcsecant