Definite Integral of Reciprocal of Root of a Squared minus x Squared

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int_0^x \frac {\d t} {\sqrt{1 - t^2} } = \arcsin x$


Proof

\(\displaystyle \int_0^x \frac {\d t} {\sqrt{1 - t^2} }\) \(=\) \(\displaystyle \sqbrk {\arcsin \frac t 1}_0^x\) Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$, Definition of Definite Integral
\(\displaystyle \) \(=\) \(\displaystyle \arcsin x - \arcsin 0\)
\(\displaystyle \) \(=\) \(\displaystyle \arcsin x\)

$\blacksquare$


Sources