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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int \frac 1 {x \sqrt {a^2} \sqrt {\paren {\frac x a}^2 - 1} } \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {\size a} \int \frac 1 {x \sqrt {\paren {\frac x a}^2 - 1} } \rd x\) |
- $\sec \theta = \dfrac x {\size a} \iff \size a \sec \theta = x$
for $\theta \in \openint 0 {\dfrac \pi 2} \cup \openint {\dfrac \pi 2} \pi$.
This substitution is valid for all $\dfrac x {\size a} \in \R \setminus \openint {-1} 1$.
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| \(\ds \paren {x > \size a}\) | \(\lor\) | \(\ds \paren {x < - \size a}\) | ||||||||||||
| \(\ds \iff \ \ \) | \(\ds \paren {\dfrac x {\size a} > 1}\) | \(\lor\) | \(\ds \paren {\dfrac x {\size a} < -1}\) |
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 |
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