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

## Theorem

$\ds \int \frac {x^3 \rd x} {\sqrt {a^2 - x^2} } = \frac {\paren {\sqrt {a^2 - x^2} }^3} 3 - a^2 \sqrt {a^2 - x^2} + C$

## Proof

With a view to expressing the problem in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

 $\ds u$ $=$ $\ds x^2$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds 2 x$ Power Rule for Derivatives

and let:

 $\ds \frac {\d v} {\d x}$ $=$ $\ds \frac x {\sqrt {a^2 - x^2} }$ $\ds \leadsto \ \$ $\ds v$ $=$ $\ds -\sqrt {a^2 - x^2}$ Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$

Then:

 $\ds \int \frac {x^3 \rd x} {\sqrt {a^2 - x^2} }$ $=$ $\ds \int x^2 \frac {x \rd x} {\sqrt {a^2 - x^2} }$ $\ds$ $=$ $\ds -x^2 \sqrt {a^2 - x^2} - \int 2 x \paren {-\sqrt {a^2 - x^2} } \rd x$ Integration by Parts $\ds$ $=$ $\ds -x^2 \sqrt {a^2 - x^2} + 2 \int x \paren {\sqrt {a^2 - x^2} } \rd x$ simplification $\ds$ $=$ $\ds -x^2 \sqrt {a^2 - x^2} + 2 \paren {-\frac {\paren {\sqrt {a^2 - x^2} }^3} 3} + C$ Primitive of $x \sqrt {a^2 - x^2}$ $\ds$ $=$ $\ds \paren {a^2 - x^2 - a^2} \sqrt {a^2 - x^2} - \frac {2 \paren {\sqrt {a^2 - x^2} }^3} 3 + C$ $\ds$ $=$ $\ds \paren {\sqrt {a^2 - x^2} }^3 - a^2 \sqrt {a^2 - x^2} - \frac {2 \paren {\sqrt {a^2 - x^2} }^3} 3 + C$ $\ds$ $=$ $\ds \frac {\paren {\sqrt {a^2 - x^2} }^3} 3 - a^2 \sqrt {a^2 - x^2} + C$

$\blacksquare$