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#.
