Primitive of x by Root of a squared minus x squared cubed

Theorem

$\ds \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x = \frac {-\paren {\sqrt {a^2 - x^2} }^5} 5 + C$

Let:

$\ds z$

$=$

$\ds a^2 - x^2$

$\ds \leadsto \ \ $

$\ds \frac {\d z} {\d x}$

$=$

$\ds -2 x$

$\ds \leadsto \ \ $

$\ds \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x$

$=$

$\ds \int \frac {z^{3/2} } {-2} \rd z$

$\ds $

$=$

$\ds \frac 1 2 \frac {-z^{5/2} } {\frac 5 2} + C$

$\ds $

$=$

$\ds \frac {-z^{5/2} } 5 + C$

simplifying

$\ds $

$=$

$\ds \frac {-\paren {\sqrt {a^2 - x^2} }^5} 5 + C$

substituting for $z$

$\blacksquare$