To get rid of the square root in the denominator, we make the trigonometric substitution x \mapsto g ( u ) with
g ( u ) = sin u / 3,
g' ( u ) = cos u / 3, and
g^{-1} ( x ) = sin^{-1} ( 3 x ),
which leads to
[ \frac{1}{81} int \frac{sin^3 u cos u}{\sqrt{1 - sin^2 u}} \ d u ]_{u = sin^{-1} ( 3 x )} =
= [ \frac{1}{81} int sin^3 u \ d u ]_{u = sin^{-1} ( 3 x )} =
= [ \frac{1}{81} int ( 1 - cos^2 u ) sin u \ d u ]_{u = sin^{-1} ( 3 x )}.
Now, we may view cos u as an inner function, with derivative - sin u, which is why this is the same as
[ - \frac{1}{81} int 1 - t^2 \ d t ]_{t = cos sin^{-1} ( 3 x ) = \sqrt{1 - 9 x^2}} =
= - \frac{1}{81} [ t - t^3 / 3 + C ]_{t = \sqrt{1 - 9 x^2}} =
= - \frac{1}{81} [ \sqrt{1 - 9 x^2} - (1 - 9 x^2)^(3 / 2) / 3 ] + C =
= - \frac{1}{243} \sqrt{1 - 9 x^2} ( 2 + 9 x^2 ) + C.