Arcsin as an Integral
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Contents
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.
- $\displaystyle \map {\arcsin_G} x = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$
$\Box$
- $\displaystyle \map \arcsin x = \map {\arcsin_A} x = \map {\arcsin_G} x = \int_0^\Theta \frac {\d x} {\sqrt {1 - x^2} }$
$\blacksquare$
