x^2 = t
int[x^3(2-x^2)^12]dx = 1/2 int[t(2-t)^12]dt
from here you need to use the chain rule
=1/2 ((t^2/2)(2-t)^12 - int[(2-t)^12]dt)
and from there just do another simple integral and replace t with x^2...
Hello, ^_^Engineer_Adam^_^!
Integrate: .∫x³(2 - x²)^{12} dx
Integrate by parts . . .
. . u .= .x² . . . . . dv .= .x(2 - x²)^{12} dx
. du .= .2x dx . . . .v .= .(-1/26)(2 - x²)^{13}
And we have: . (-1/26)x²(2 - x²)^{13} + (1/13) ∫ x(2 - x²)^{13} dx
. . Finally let: u .= .2 - x² . . . etc.
Well you have been answered but i will show you my way in detailed :
int[x^3(2-x^2)^12]dx
t=x^2
dt = 2xdx ==> xdx = dt/2
int[x^3(2-x^2)^12]dx = int[x^2(2-x^2)^12]xdx = (xdx = dt/2 x^2=t) =
= int[t(2-t)^12]1/2dt = 1/2int[t(2-t)^12]dt
now integration by parts : int[f(t)g`(t)]dt = f(t)g(t) - int[f`(t)g(t)]dt
let f(t) = t , g(t) = ((2-t)^13)/13
f`(t) = 1 , g`(t) = (2-t)^12
then :
int[t(2-t)^12]dt = t((2-t)^13)/13 - 1/13int[(2-t)^13]dt
from here you can go by yourself...