How do you find the integral intx^3*sqrt(9-x^2)dx ?

1 Answer
Aug 29, 2014

=(9-x^2)^(5/2)/5-3(9-x^2)^(3/2)+c, where c is a constant

Explanation :

=intx^3*sqrt(9-x^2)dx

Using Integration by Substitution,

let's assume 9-x^2=t^2, then

-2xdx=2tdt =>xdx=-tdt

=int-(9-t^2)t^2dt

=int(t^4-9t^2)dt

=intt^4dt-9intt^2dt

=t^5/5-9t^3/3+c, where c is a constant

=t^5/5-3t^3+c, where c is a constant

Substituting t back,

=(9-x^2)^(5/2)/5-3(9-x^2)^(3/2)+c, where c is a constant