Primitive of Reciprocal of x by Root of x squared minus a squared/Arccosine Form

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Theorem

$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \arccos \size {\frac a x} + C$

for $0 < a < \size x$.


Proof

We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:

$x > a$

or:

$x < -a$

where it is assumed that $a > 0$.


Hence:

\(\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} }\) \(=\) \(\ds \frac 1 a \arcsec \size {\frac x a} + C\) Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$: Arcsecant Form
\(\ds \) \(=\) \(\ds \frac 1 a \arccos \size {\frac a x} + C\) Arcsecant of Reciprocal equals Arccosine

$\blacksquare$


Also see


Sources

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