Integral Resulting in Arcsecant

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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

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

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:

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

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


Thus:

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

and so:

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


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 $(A)$:

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


If $(B)$:

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

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