Substitute #x=sint#, #dx=costdt#. As the integrand is defined only for #x in (-1,1)# then #t# will vary accordingly in #(-pi/2,pi/2)# and we must keep in mind that in this interval #cost# is positive.
So:
#int x^3/(sqrt(1-x^2) )dx = int sin^3t/sqrt(1-sin^2t) cost dt#
#int x^3/(sqrt(1-x^2) )dx = int sin^3t/cost cost dt#
#int x^3/(sqrt(1-x^2) )dx = int sin^3tdt#
Now:
#int sin^3tdt = int sin^2t sint dt#
#int sin^3tdt = -int (1-cos^2t) d(cost)#
and using the linearity of the integral:
#int sin^3tdt = int cos^2t d(cost) - int d(cost)#
#int sin^3tdt = cos^3t/3 - cost+C#
to undo the substitution note that:
#cost = sqrt(1-sin^2t) = sqrt(1-x^2)#
So:
#int x^3/(sqrt(1-x^2) )dx = 1/3(1-x^2)^(3/2) - sqrt (1-x^2) +C#
#int x^3/(sqrt(1-x^2) )dx = sqrt(1-x^2)(1/3 -x^2/3 - 1)+C#
#int x^3/(sqrt(1-x^2) )dx = -sqrt(1-x^2)((x^2+2)/3)+C#
It is however probably more straightforward not to use a trigonometric substitution, in fact:
#int x^3/sqrt(1-x^2) dx = int x^2 (2xdx)/(2sqrt(1-x^2))#
and we can integrate by parts:
#int x^3/sqrt(1-x^2) dx = -int x^2 d(sqrt(1-x^2))#
#int x^3/sqrt(1-x^2) dx = -x^2(sqrt(1-x^2)) + 2int xsqrt(1-x^2)dx#
and finally:
#int x^3/sqrt(1-x^2) dx = -x^2(sqrt(1-x^2)) - int sqrt(1-x^2)d(1-x^2)#
#int x^3/sqrt(1-x^2) dx = -x^2(sqrt(1-x^2)) - 2/3 (1-x^2)^(3/2)+C#
#int x^3/sqrt(1-x^2) dx = sqrt(1-x^2) (-x^2 - 2/3 +2/3x^2)+C#
#int x^3/sqrt(1-x^2) dx = -sqrt(1-x^2) ((x^2 + 2)/3) +C#
