Thread: Help with integrals - chain rule

Integrate:

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...

it should be like that after integrating in parts :

1/2( (t(2-t)^13)/13 - int[(2-t)^12]dt)

Originally Posted by Tankado

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...

but theres an x^3?... how did t = x^3?
and the method is diff from the method of my book
and thenks for replying tankado

Rewite as:
(x^2(2-x^2))(xdx)

u=2-x^2, x^2=2-u, du=-2xdx.

(-1/2)(1-u)(u)^12 du

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...

Not necessary to use integration by parts, just set u = 2 - x², and it becomes very easy.

So there is a way more easier?...No wonder i didnt understand the previuos posts...thats becuase im still taking up basic in integrals...( havent read about int by parts yet)
could you please show to me the solution
When u = 2-x^2
then what is x^3?

What I'm trying to say is