Arcsin as an Integral

Theorem

$\displaystyle \map \arcsin x = \int_0^\Theta \frac {\d x} {\sqrt {1 - x^2} }$

Proof

Lemma 1

Let $\sin_A$ be the analytic sine function for real numbers.

Let $\arcsin_A$ denote the real arcsine function.

Then:

$\ds \map {\arcsin_A} x = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$

$\Box$

Lemma 2

Let, $\sin_G$ be the Geometric Sine from Definition:Sine/Definition from Circle. $\arcsin_G$ is the inverse of this function.

$\displaystyle \map {\arcsin_G} x = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$

$\Box$

$\displaystyle \map \arcsin x = \map {\arcsin_A} x = \map {\arcsin_G} x = \int_0^\Theta \frac {\d x} {\sqrt {1 - x^2} }$

$\blacksquare$