Primitive of Arcsecant of x over a/Formulation 2

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Theorem

$\ds \int \arcsec \frac x a \rd x = x \arcsec \frac x a - a \ln \size {x + \sqrt {x^2 - a^2} } + C$

for $x^2 > 1$.

$\arcsec \dfrac x a$ is undefined on the real numbers for $x^2 < 1$.


Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \arcsec \frac x a\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \dfrac a {\size x \sqrt {x^2 - a^2} }\) Derivative of $\arcsec \dfrac x a$


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds x\) Primitive of Constant


We then have:

\(\ds \int \arcsec \frac x a \rd x\) \(=\) \(\ds x \arcsec \frac x a - \int x \paren {\dfrac a {\size x \sqrt {x^2 - a^2} } } \rd x + C\) Integration by Parts
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds x \arcsec \frac x a - \int \frac x {\size x} \paren {\frac a {\sqrt {x^2 - a^2} } } \rd x + C\) rearrangement


Let $x > 1$.

Then:

\(\ds \int \frac x {\size x} \paren {\frac a {\sqrt {x^2 - a^2} } } \rd x\) \(=\) \(\ds \int \frac a { {\sqrt {x^2 - a^2} } } \rd x\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds a \int \frac 1 { {\sqrt {x^2 - a^2} } } \rd x\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds a \size {\ln \size {x + \sqrt {x^2 - a^2} } } + C\) Integral of One Over Square Root of Binomial
\(\ds \) \(=\) \(\ds a \ln \size {x + \sqrt {x^2 - a^2} } + C\) as argument of logarithm is positive


Similarly, let $x < -1$.

Then:

\(\ds \int \frac x {\size x} \paren {\frac a {\sqrt {x^2 - a^2} } } \rd x\) \(=\) \(\ds \int \paren {-1} \frac a { {\sqrt{x^2 - a^2} } } \rd x\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds -a \int \frac 1 { {\sqrt{x^2 - a^2} } } \rd x\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds -a \size {\ln \size {x + \sqrt {x^2 - a^2} } } + C\) Integral of One Over Square Root of Binomial
\(\ds \) \(=\) \(\ds -a \paren {-\ln \size {x + \sqrt {x^2 - a^2} } } + C\) as argument of logarithm is negative
\(\ds \) \(=\) \(\ds a \ln \size {x + \sqrt {x^2 - a^2} } + C\) as argument of logarithm is negative

$\blacksquare$


Sources

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