Arcsine as Integral/Lemma 1
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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:
- $x = \sin_A \theta \iff \theta = \map {\arcsin_A} x$
Then:
\(\ds \d x\) | \(=\) | \(\ds \cos_A \theta \rd \theta\) | Derivative of Sine Function | |||||||||||
\(\ds \int \frac {\d x} {\sqrt {1 - x^2} }\) | \(=\) | \(\ds \int \frac {\cos_A \theta \rd \theta } {\sqrt {1 - \sin_A^2 \theta} }\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\cos_A \theta \rd \theta } {\sqrt {\cos_A^2 \theta} }\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \int 1 \rd \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \theta + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\arcsin_A} x + C\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int_0^x \frac {\d x} {\sqrt {1 - x^2} }\) | \(=\) | \(\ds \map {\arcsin_A} x\) | Fundamental Theorem of Calculus: Second Part |
$\blacksquare$