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

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

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.33$: Weierstrass ($1815$ – $1897$)