# Primitive of Square of Arcsine of x over a

## Theorem

$\displaystyle \int \left({\arcsin \frac x a}\right)^2 \ \mathrm d x = x \left({\arcsin \frac x a}\right)^2 - 2 x + 2 \sqrt{a^2 - x^2} \arcsin \frac x a + C$

## Proof

Let:

 $\displaystyle u$ $=$ $\displaystyle \arcsin \frac x a$ $(1):\quad$ $\displaystyle \implies \ \$ $\displaystyle \sin u$ $=$ $\displaystyle \frac x a$ Definition of Arcsine $(2):\quad$ $\displaystyle \implies \ \$ $\displaystyle \cos u$ $=$ $\displaystyle \sqrt {1 - \frac {x^2} {a^2} }$ Sum of Squares of Sine and Cosine

Then:

 $\displaystyle \int \left({\arcsin \frac x a}\right)^2 \ \mathrm d x$ $=$ $\displaystyle a \int u^2 \cos u \ \mathrm d x$ Primitive of Function of Arcsine $\displaystyle$ $=$ $\displaystyle a \left({2 u \cos u + \left({u^2 - 2}\right) \sin u}\right) + C$ Primitive of $x^2 \cos a x$ $\displaystyle$ $=$ $\displaystyle a \left({2 \arcsin \frac x a \sqrt {1 - \frac {x^2} {a^2} } + \left({\left({\arcsin \frac x a}\right)^2 - 2}\right) \frac x a}\right) + C$ substituting for $u$, $\sin u$ and $\cos u$ $\displaystyle$ $=$ $\displaystyle x \left({\arcsin \frac x a}\right)^2 - 2 x + 2 \sqrt{a^2 - x^2} \arcsin \frac x a + C$ simplifying

$\blacksquare$