Question

integral of x^3/(sqrt(16-x^2))

Answer 1

let u = 16 - x^2 ---------> x^2 = (16-u)
du = -2x dx or
-1/2 du = x dx

see ur integral becomes
int (-1/2 (16-u))/sqrt(u) du
= int -8/sqrt(u) du + int (1/2 u)/sqrt(u) du
= int -8(u)^-1/2 du + int 1/2 u^1/2 du
now, integrate this

Answer 2

use the basic formula :