I am having trouble with the Schaum Mathematical Handbook of Formulas and Tables (1968, Murray R. Spiegel) item $14.246$:
$$\int x^2 \sqrt {a^2 - x^2} \, \mathrm d x = -\frac {x (\sqrt {a^2 - x^2})^3} 4 + \frac {a^2 x \sqrt {a^2 - x^2} } 8 + \frac {a^4} 8 \arcsin \frac x a + C$$
I attacked this by substituting $z = x^2$ which led me to:
$$\int x^2 \sqrt {a^2 - x^2} \, \mathrm d x = -\frac 1 2 \int \sqrt {z (a^2 - z)} \, \mathrm d z$$
There is a standard result here (14.122 in Schaum):
$$\int \sqrt {(a x + b) (p x + q)} \, \mathrm d x = \frac {2 a p x + b p + a q} {4 a p} \sqrt {(a x + b) (p x + q)} - \frac {(b p - a q)^2} {8 a p} \int \frac {\mathrm d x} {\sqrt {(a x + b) (p x + q)} }$$
Setting $a \gets 1$, $b \gets 0$, $p \gets -1$, $q \gets a^2$, $x \gets z$ leads us to:
$$\frac 1 2 \int \sqrt {z (a^2 - z)} \, \mathrm d z= \frac {2 z - a^2} 8 \sqrt {z (a^2 - z)} + \frac {a^4} {16} \int \frac {\mathrm d z} {\sqrt {z (a^2 - z)} }$$
Then I went to Handbook of Mathematical Functions (1964: Milton Abramowitz and Irene A. Stegun) to use item $3.3.27$:
$$\int \frac {\mathrm d x} {\sqrt {(a x + b) (p x + q)} } = \dfrac {-1} {\sqrt {-a p} } \arcsin \left( {\dfrac {2 a p x + b p + a q} {a q - b p} }\right) + C$$
which is appropriate where $a > 0$ and $p < 0$, as they are here: $a = 1$, $p = -1$.
using the same substitutions as above, this gives us:
$$\frac {2 z - a^2} 8 \sqrt {z (a^2 - z)} + \frac {a^4} {16} \int \frac {\mathrm d z} {\sqrt {z (a^2 - z)} } = \frac {2 z - a^2} 8 \sqrt {z (a^2 - z)} + \frac {a^4} {16} \left( {-\arcsin \frac {2 z - a^2} 4}\right) + C$$
and substituting back for $z$:
$$= \frac {2 x^2 - a^2} 8 \sqrt {x^2 (a^2 - x^2)} + \frac {a^4} {16} \left( {-\arcsin \frac {2 x^2 - a^2} 4}\right) + C$$
All well and good with the left hand term, this easily equates to $-\dfrac {x (\sqrt {a^2 - x^2})^3} 4 + \dfrac {a^2 x \sqrt {a^2 - x^2} } 8$
But I'm stumped as to how to get to $\dfrac {a^4} 8 \arcsin \dfrac x a$ with that last term.
Any pointers? Many thanks.
EDIT: While it is true that Find a primitive of $x^2\sqrt{a^2 - x^2}$ answers the solution of this integral, I am specifically interested in the solution which uses the given substitution and standard integrals.
