Primitive of x squared by Root of a squared minus x squared/Also presented as

Primitive of $x^2 \sqrt {a^2 - x^2}$: Also presented as

This result is also seen presented in the form:

$\ds \int x^2 \sqrt {a^2 - x^2} \rd x = \frac {a^4} 8 \arcsin \frac x a - \frac {x \sqrt {a^2 - x^2} \paren {a^2 - 2 x^2} } 8 + C$

$\ds \int x^2 \sqrt {a^2 - x^2} \rd x$

$=$

$\ds -\frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {a^2 x \sqrt {a^2 - x^2} } 8 + \frac {a^4} 8 \arcsin \frac x a + C$

$\ds $

$=$

$\ds \frac {a^4} 8 \arcsin \frac x a + \frac {a^2 x \sqrt {a^2 - x^2} } 8 - \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + C$

reversing the order for convenience

$\ds $

$=$

$\ds \frac {a^4} 8 \arcsin \frac x a + \frac {a^2 x \sqrt {a^2 - x^2} - 2 x \paren {\sqrt {a^2 - x^2} }^3} 8 + C$

$\ds $

$=$

$\ds \frac {a^4} 8 \arcsin \frac x a + \frac {x \sqrt {a^2 - x^2} \paren {a^2 - 2 \paren {\sqrt {a^2 - x^2} }^2} } 8 + C$

$\ds $

$=$

$\ds \frac {a^4} 8 \arcsin \frac x a + \frac {x \sqrt {a^2 - x^2} \paren {a^2 - 2 \paren {a^2 - x^2} } } 8 + C$

simplifying

$\ds $

$=$

$\ds \frac {a^4} 8 \arcsin \frac x a - \frac {x \sqrt {a^2 - x^2} \paren {a^2 - 2 x^2} } 8 + C$

arranging into required form

$\blacksquare$