Integral of Arcsine Function

Theorem

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

for $x \in [-1..1]$.

Proof

 * $\displaystyle \int \arcsin {x} \ \mathrm d x = \int 1 \cdot \arcsin {x} \ \mathrm d x$

Using: we obtain:
 * Integration by Parts
 * Derivative of Arcsine Function
 * Integration of a Constant


 * $\displaystyle \int \arcsin {x} \ \mathrm d x = x \arcsin x - \int \dfrac x {\sqrt {1 - x^2}} \ \mathrm dx$

Substitute:


 * $u = 1-x^2$
 * $\mathrm du = -2x \ \mathrm dx$
 * $\implies -\dfrac 1 2 \mathrm du = x \ \mathrm dx$

which yields: