Primitive of Function of Arcsecant

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int \map F {\arcsec \frac x a} \rd x = a \int \map F u \sec u \tan u \rd u$

where $u = \arcsec \dfrac x a$.


Proof

First note that:

\(\displaystyle u\) \(=\) \(\displaystyle \arcsec \frac x a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle a \sec u\) Definition of Arcsecant


Then:

\(\displaystyle u\) \(=\) \(\displaystyle \arcsec \frac x a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle \frac a {\size x {\sqrt {x^2 - a^2} } }\) Derivative of Arcsecant Function: Corollary 1
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int \map F {\arcsec \frac x a} \rd x\) \(=\) \(\displaystyle \int \map F u \, \frac {\size x {\sqrt {x^2 - a^2} } } a \rd u\) Primitive of Composite Function
\(\displaystyle \) \(=\) \(\displaystyle \int \map F u \, \frac {\size {a \sec u} {\sqrt {a^2 \sec^2 u - a^2} } } a \rd u\) Definition of $x$
\(\displaystyle \) \(=\) \(\displaystyle \int \map F u \, \size {\sec u} {\sqrt {a^2 \sec^2 u - a^2} } \rd u\)
\(\displaystyle \) \(=\) \(\displaystyle \int \map F u \, a \sec u {\sqrt {\sec^2 u - 1} } \rd u\) $\sec u > 0$ in this domain
\(\displaystyle \) \(=\) \(\displaystyle \int \map F u \, a \sec u \tan u \rd u\) Difference of Squares of Secant and Tangent
\(\displaystyle \) \(=\) \(\displaystyle a \int \map F u \sec u \tan u \rd u\) Primitive of Constant Multiple of Function

$\blacksquare$


Also see