Primitive of Function of Arcsecant
Jump to navigation
Jump to search
Theorem
- $\ds \int \map F {\arcsec \frac x a} \rd x = a \int \map F u \sec u \tan u \rd u$
where $u = \arcsec \dfrac x a$.
Proof
First note that:
| \(\ds u\) | \(=\) | \(\ds \arcsec \frac x a\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds a \sec u\) | Definition of Arcsecant |
Then:
| \(\ds u\) | \(=\) | \(\ds \arcsec \frac x a\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac a {\size x {\sqrt {x^2 - a^2} } }\) | Derivative of Arcsecant Function: Corollary 1 | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int \map F {\arcsec \frac x a} \rd x\) | \(=\) | \(\ds \int \map F u \, \frac {\size x {\sqrt {x^2 - a^2} } } a \rd u\) | Primitive of Composite Function | ||||||||||
| \(\ds \) | \(=\) | \(\ds \int \map F u \frac {\size {a \sec u} {\sqrt {a^2 \sec^2 u - a^2} } } a \rd u\) | Definition of $x$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int \map F u \size {\sec u} {\sqrt {a^2 \sec^2 u - a^2} } \rd u\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int \map F u a \sec u {\sqrt {\sec^2 u - 1} } \rd u\) | $\sec u > 0$ in this domain | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int \map F u a \sec u \tan u \rd u\) | Difference of Squares of Secant and Tangent | |||||||||||
| \(\ds \) | \(=\) | \(\ds a \int \map F u \sec u \tan u \rd u\) | Primitive of Constant Multiple of Function |
$\blacksquare$