# Need help with indefinite integral

• Mar 16th 2010, 02:36 PM
blackwrx
Need help with indefinite integral
I am pretty sure I need to use integration by parts, but I cannot get anywhere with this.

$\int 2x^3\cos(x^2)\,dx$
• Mar 16th 2010, 02:52 PM
tonio
Quote:

Originally Posted by blackwrx
I am pretty sure I need to use integration by parts, but I cannot get anywhere with this.

$\int 2x^3\cos(x^2)\,dx$

Indeed. You can try $u=x^2\,,\,u'=2x\,,\,\,v'=2x\cos x\,,\,v=\sin x^2$ , so

$\int 2x^3\cos(x^2)\,dx=x^2\sin x^2-\int 2x\sin x^2\,dx$ $= x^2\sin x^2+\cos x^2+C$ , the last integral being (almost) immediate

since it is of the form $\int f'(x)\sin f(x),dx=-\cos f(x)+C$

Tonio
• Mar 16th 2010, 02:53 PM
Moo
Hello,

First substitute $u=x^2$ and then use once an integration by parts.
• Mar 16th 2010, 03:00 PM
Aryth
You could try a substitution first...

Let $w = x^2$, $dw = 2x ~dx$

$\int wcos(w) ~dw$

That is easily integrable by parts.

$u = w$, $du = 1 ~dw$, $dv = cos(w) ~dw$, $v = sin(w)$

$= wsin(w) - \int sin(w) ~dw$

$= wsin(w) + cos(w) + C$

$= x^2sin(x^2) + cos(x^2) + C$

And there you go.
• Mar 16th 2010, 03:21 PM
blackwrx
Where did the $2xcos(x)$ come from? What happened to the original $2x^3$?
• Mar 16th 2010, 03:23 PM
blackwrx
Nevermind, I see with the substitution that dw=2x times the w=x^2 does indeed solve my problem. Thanks guys.