# Integral Resulting in Arcsecant

 It has been suggested that this page or section be merged into Primitive of Reciprocal of x by Root of x squared minus a squared. (Discuss)

## 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

 $\ds \int \frac 1 {x \sqrt{x^2 - a^2} } \ \mathrm dx$ $=$ $\ds \int \frac 1 {x \sqrt{a^2 \left({\frac {x^2}{a^2} - 1}\right)} } \ \mathrm dx$ factor $a^2$ out of the radicand $\ds$ $=$ $\ds \int \frac 1 {x \sqrt{a^2} \sqrt{\left({\frac x a}\right)^2 - 1} } \ \mathrm dx$ $\ds$ $=$ $\ds \frac 1 {\left \vert {a} \right \vert} \int \frac 1 {x \sqrt{\left({\frac x a}\right)^2 - 1} } \ \mathrm dx$
$\sec \theta = \dfrac x {\left\vert{a}\right\vert} \iff \left\vert{a}\right\vert \sec \theta = x$

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

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

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

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

Thus:

 $\ds \left \vert{a}\right \vert \sec \theta$ $=$ $\ds x$ from above $\ds \implies \ \$ $\ds \left \vert {a}\right \vert \sec \theta \tan \theta \frac {\mathrm d \theta}{\mathrm dx}$ $=$ $\ds 1$ differentiate WRT $x$, Derivative of Secant Function, Chain Rule for Derivatives

and so:

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

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 $(A)$:

 $\ds \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta} {\left \vert \tan \theta \right \vert} \ \mathrm d \theta$ $=$ $\ds \frac 1 {\left \vert {a} \right \vert} \int \mathrm d \theta$ definition of absolute value $\ds$ $=$ $\ds \frac 1 {\left \vert {a} \right \vert} \theta + C$ Integral of Constant $\ds$ $=$ $\ds \frac 1 {\left \vert {a} \right \vert} \operatorname{arcsec} \frac x {\left \vert {a} \right \vert} + C$ definition of arcsecant

If $(B)$:

 $\ds \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta} {\left \vert \tan \theta \right \vert} \ \mathrm d \theta$ $=$ $\ds \frac 1 {\left \vert {a} \right \vert} \int -1 \ \mathrm d \theta$ definition of absolute value $\ds$ $=$ $\ds -\frac 1 {\left \vert {a} \right \vert} \theta + C$ Integral of Constant $\ds$ $=$ $\ds -\frac 1 {\left \vert {a} \right \vert} \operatorname{arcsec} \frac x {\left \vert {a} \right \vert} + C$ definition of arcsecant