Definite Integral of Reciprocal of Root of a Squared minus x Squared
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Theorem
- $\displaystyle \int_0^x \frac {\d t} {\sqrt{1 - t^2} } = \arcsin x$
Proof
\(\ds \int_0^x \frac {\d t} {\sqrt{1 - t^2} }\) | \(=\) | \(\ds \intlimits {\arcsin \frac t 1} 0 x\) | Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$, Definition of Definite Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \arcsin x - \arcsin 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \arcsin x\) |
$\blacksquare$
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.33$: Weierstrass ($\text {1815}$ – $\text {1897}$)