# Arcsin as an Integral/Lemma 1

## Lemma

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

Let $\arcsin_A$ denote the real arcsine function.

Then:

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

## Proof

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:

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

Then:

 $\displaystyle \d x$ $=$ $\displaystyle \cos_A \theta \rd \theta$ Derivative of Sine Function $\displaystyle \int \frac {\d x} {\sqrt {1 - x^2} }$ $=$ $\displaystyle \int \frac {\d x} {\cos_A \theta} \cos_A \theta \rd \theta$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \int 1 \rd \theta$ $\displaystyle$ $=$ $\displaystyle \theta + C$ $\displaystyle$ $=$ $\displaystyle \map {\arcsin_A} x + C$ $\displaystyle \leadsto \ \$ $\displaystyle \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$ $=$ $\displaystyle \map {\arcsin_A} x$ Fundamental Theorem of Calculus: Second Part

$\Box$