Integral of Arcsine Function

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Theorem

$\displaystyle \int \arcsin x \rd x = x \arcsin x + \sqrt {1 - x^2} + C$

for $x \in \left[{-1 \,.\,.\, 1}\right]$.


Proof

$\displaystyle \int \arcsin x \rd x = \int 1 \cdot \arcsin x \rd x$


Using:

we obtain:

$\displaystyle \int \arcsin x \rd x = x \arcsin x - \int \dfrac x {\sqrt {1 - x^2}} \rd x$


Substitute:

\(\displaystyle u\) \(=\) \(\displaystyle 1 - x^2\)
\(\displaystyle \dfrac {\d u} {\d x}\) \(=\) \(\displaystyle -2 x\) differentiate WRT $x$
\(\displaystyle \implies \ \ \) \(\displaystyle -\frac 1 2 \frac {\d u} {\d x}\) \(=\) \(\displaystyle x\)
\(\displaystyle \int \arcsin x \rd x\) \(=\) \(\displaystyle x \arcsin x - \int \frac {- \frac 1 2} {\sqrt u} \frac {\d u} {\d x} \rd x\)
\(\displaystyle \int \arcsin x \rd x\) \(=\) \(\displaystyle x \arcsin x + \frac 1 2 \int u^{-1/2} \rd u\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle x \arcsin x + u^{1/2} + C\) Integral of Power
\(\displaystyle \) \(=\) \(\displaystyle x \arcsin x + \sqrt {1-x^2} + C\) from $u = 1 - x^2$

$\blacksquare$


Sources