Integral Resulting in Arcsecant

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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
\(\ds \) \(=\) \(\ds \int \frac 1 {x \sqrt {a^2} \sqrt {\paren {\frac x a}^2 - 1} } \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 {\size a} \int \frac 1 {x \sqrt {\paren {\frac x a}^2 - 1} } \rd x\)

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:

\(\ds \paren {x > \size a}\) \(\lor\) \(\ds \paren {x < - \size a}\)
\(\ds \iff \ \ \) \(\ds \paren {\dfrac x {\size a} > 1}\) \(\lor\) \(\ds \paren {\dfrac x {\size a} < -1}\)

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


Thus:

\(\ds \size a \sec \theta\) \(=\) \(\ds x\) from above
\(\ds \implies \ \ \) \(\ds \size a \sec \theta \tan \theta \frac {\d \theta} {\d x}\) \(=\) \(\ds 1\) 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

Also see