Arcsin as an Integral

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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, the one defined by Definition:Sine/Real Numbers. $\arcsin_A$ is the inverse of this function.

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

$\Box$


Lemma 2

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

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

$\Box$


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

$\blacksquare$