Question

How do you find the integral `intx^3/(sqrt(16-x^2))dx` ?

Answer 1

`=(16-x^2)^(3/2)/3-16sqrt(16-x^2)+c`, where `c` is a constant

Solution

`=intx^3/sqrt(16-x^2)dx=int(x^2*x)/sqrt(16-x^2)dx`

Using Integration by Substitution

let's assume `(16-x^2)=t^2`, `=>-2xdx=2tdt`

`xdx=-tdt`

`=int(-(16-t^2)t)/tdt`

`=int(t^2-16)dt`

`=intt^2dt-16intdt`

`=t^3/3-16t+c`, where `c` is a constant

`=(16-x^2)^(3/2)/3-16sqrt(16-x^2)+c`, where `c` is a constant