# Integral of Arcsine Function

 It has been suggested that this page or section be merged into Primitive of Arcsine of x over a. (Discuss)

## 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:

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

$\blacksquare$