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 $\dfrac x {\size a} \in \R \setminus \openint {-1} 1$.

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By hypothesis:

\(\ds \paren {x > \size a}\)

\(\lor\)

\(\ds \paren {x < - \size a}\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \paren {\dfrac x {\size a} > 1}\)

\(\lor\)

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

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

Thus:

\(\ds \size a \sec \theta\)

\(=\)

\(\ds x\)

from above

\(\ds \leadsto \ \ \)

\(\ds \size a \sec \theta \tan \theta \frac {\d \theta} {\d x}\)

\(=\)

\(\ds 1\)

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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Also see

• Derivative of Arcsecant Function