Primitive of Square of Arcsine of x over a

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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

$\blacksquare$


Also see


Sources