# Primitive of x squared by Arccosine of x over a

## Theorem

$\displaystyle \int x^2 \arccos \frac x a \rd x = \frac {x^3} 3 \arccos \frac x a - \frac {\paren {x^2 + 2 a^2} \sqrt {a^2 - x^2} } 9 + C$

## Proof

With a view to expressing the primitive in the form:

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

let:

 $\displaystyle u$ $=$ $\displaystyle \arccos \frac x a$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d u} {\d x}$ $=$ $\displaystyle \frac {-1} {\sqrt {a^2 - x^2} }$ Derivative of $\arccos \dfrac x a$

and let:

 $\displaystyle \frac {\d v} {\d x}$ $=$ $\displaystyle x^2$ $\displaystyle \leadsto \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {x^3} 3$ Primitive of Power

Then:

 $\displaystyle \int x^2 \arccos \frac x a \rd x$ $=$ $\displaystyle \frac {x^3} 3 \arccos \frac x a - \int \frac {x^3} 3 \paren {\frac {-1} {\sqrt {a^2 - x^2} } } \rd x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {x^3} 3 \arccos \frac x a + \frac 1 3 \int \frac {x^3 \rd x} {\sqrt {a^2 - x^2} } + C$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac {x^3} 3 \arccos \frac x a + \frac 1 3 \paren {\frac {\paren {\sqrt {a^2 - x^2} }^3} 3 - a^2 \sqrt {a^2 - x^2} } + C$ Primitive of $\dfrac {x^3} {\sqrt {a^2 - x^2} }$ $\displaystyle$ $=$ $\displaystyle \frac {x^3} 3 \arccos \frac x a - \frac {\paren {x^2 + 2 a^2} \sqrt {a^2 - x^2} } 9 + C$ simplifying

$\blacksquare$