Theorem: Integral Resulting in Arcsecant

:<nowiki>$\ds \int \frac 1 {x \sqrt {x^2 - a^2} } \rd x = \begin{cases}
\dfrac 1 {\size a} \arcsec \dfrac x {\size a} + C & : x > \size a \\
-\dfrac 1 {\size a} \arcsec \dfrac x {\size a} + C & : x < -\size a
\end{cases}$</nowiki>
where $a$ is a constant.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac 1 {x \sqrt {x^2 - a^2} } \rd x
      | r = \int \frac 1 {x \sqrt {a^2 \paren {\frac {x^2} {a^2} - 1} } } \rd x
      | c = factor $a^2$ out of the radicand
}}
{{eqn | r = \int \frac 1 {x \sqrt {a^2} \sqrt {\paren {\frac x a}^2 - 1} } \rd x
      | c = 
}}
{{eqn | r = \frac 1 {\size a} \int \frac 1 {x \sqrt {\paren {\frac x a}^2 - 1} } \rd x
}}
{{end-eqn}}
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 $\dfrac x {\size a} \in \R \setminus \openint {-1} 1$.
{{MissingLinks}}
By hypothesis:
{{begin-eqn}}
{{eqn | l = \paren {x > \size a}
      | o = \lor
      | r = \paren {x < - \size a}
      | c =
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {\dfrac x {\size a} > 1}
      | o = \lor
      | r = \paren {\dfrac x {\size a} < -1}
      | c =
}}
{{end-eqn}}
so this substitution will not change the domain of the integrand.
Thus:
{{begin-eqn}}
{{eqn | l = \size a \sec \theta 
      | r = x
      | c = from above
}}
{{eqn | ll= \leadsto
      | l = \size a \sec \theta \tan \theta \frac {\d \theta} {\d x}
      | r = 1
      | c = Differentiate {{WRT|Differentiation}} $x$, Derivative of Secant Function, Chain Rule for Derivatives
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \int \frac 1 {x\sqrt {x^2 - a^2} } \rd x
      | r = \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 
      | c = from above
}}
{{eqn | r = \frac 1 {\size a} \int \frac {\tan \theta} {\sqrt {\sec^2 \theta - 1} } \rd \theta
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {\size a} \int \frac {\tan \theta} {\sqrt {\tan^2 \theta} \rd \theta}
      | c = {{Corollary|Sum of Squares of Sine and Cosine|1}}
}}
{{eqn | r = \frac 1 {\size a} \int \frac {\tan \theta} {\size {\tan \theta} } \rd \theta
      | c = 
}}
{{end-eqn}}
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 $(A)$:
{{begin-eqn}}
{{eqn | l = \frac 1 {\size a} \int \frac {\tan \theta} {\size {\tan \theta} } \rd \theta
      | r = \frac 1 {\size a} \int \rd \theta
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = \frac 1 {\size a} \theta + C
      | c = Integral of Constant
}}
{{eqn | r = \frac 1 {\size a} \arcsec \frac x {\size a} + C
      | c = {{Defof|Arcsecant}}
}}
{{end-eqn}}
If $(B)$:
{{begin-eqn}}
{{eqn | l = \frac 1 {\size a} \int \frac {\tan \theta} {\size {\tan \theta} } \rd \theta
      | r = \frac 1 {\size a} \int -1 \rd \theta
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = -\frac 1 {\size a} \theta + C
      | c = Integral of Constant
}}
{{eqn | r = -\frac 1 {\size a} \arcsec \frac x {\size a} + C
      | c = {{Defof|Arcsecant}}
}}
{{end-eqn}}
{{proofread}}