Arcsin as an Integral/Lemma 1

From ProofWiki
Jump to navigation Jump to search

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

$\Box$