Primitive of Arcsecant of x over a
(Redirected from Integral of Arcsecant Function)
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Theorem
Formulation 1
- $\ds \int \arcsec \frac x a \rd x = \begin {cases}
x \arcsec \dfrac x a - a \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ x \arcsec \dfrac x a + a \map \ln {x + \sqrt {x^2 - a^2} } + C & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\ \end {cases}$
Formulation 2
- $\ds \int \arcsec \frac x a \rd x = x \arcsec \frac x a - a \ln \size {x + \sqrt {x^2 - a^2} } + C$
for $x^2 > 1$.
$\arcsec \dfrac x a$ is undefined on the real numbers for $x^2 < 1$.