Arcsine as Integral/Lemma 1

Lemma

Let $\arcsin_A$ denote the real arcsine function.

Then:

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

For this proof only, let $\sin_A$ be the analytic sine function.

Consider:

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

Let:

$x = \sin_A \theta \iff \theta = \map {\arcsin_A} x$

Then:

$\ds \d x$

$=$

$\ds \cos_A \theta \rd \theta$

$\ds \int \frac {\d x} {\sqrt {1 - x^2} }$

$=$

$\ds \int \frac {\cos_A \theta \rd \theta } {\sqrt {1 - \sin_A^2 \theta} }$

$\ds $

$=$

$\ds \int \frac {\cos_A \theta \rd \theta } {\sqrt {\cos_A^2 \theta} }$

$\ds $

$=$

$\ds \int 1 \rd \theta$

$\ds $

$=$

$\ds \theta + C$

$\ds $

$=$

$\ds \map {\arcsin_A} x + C$

$\ds \leadsto \ \ $

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

$=$

$\ds \map {\arcsin_A} x$

$\blacksquare$