# Primitive of x squared over Root of a squared minus x squared

## Theorem

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

## Proof

With a view to expressing the problem in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \mathrm d x$

let:

 $\displaystyle u$ $=$ $\displaystyle x$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d u}{\mathrm d x}$ $=$ $\displaystyle 1$ Power Rule for Derivatives

and let:

 $\displaystyle \frac {\mathrm d v}{\mathrm d x}$ $=$ $\displaystyle \frac x {\sqrt {a^2 - x^2} }$ $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle -\sqrt {a^2 - x^2}$ Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$

Then:

 $\displaystyle \int \frac {x^2 \ \mathrm d x} {\sqrt {a^2 - x^2} }$ $=$ $\displaystyle \int x \frac {x \ \mathrm d x} {\sqrt {a^2 - x^2} }$ $\displaystyle$ $=$ $\displaystyle -x \sqrt {a^2 - x^2} - \int \left({-\sqrt {a^2 - x^2} }\right) \ \mathrm d x$ Integration by Parts $\displaystyle$ $=$ $\displaystyle -x \sqrt {a^2 - x^2} + \int \left({\sqrt {a^2 - x^2} }\right) \ \mathrm d x$ simplifying $\displaystyle$ $=$ $\displaystyle -x \sqrt {a^2 - x^2} + \left({\frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a}\right) + C$ Primitive of $\sqrt {a^2 - x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C$

$\blacksquare$