Question

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

Answer 1

`=(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