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

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

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

$\displaystyle \int 2x^3\cos(x^2)\,dx$ - Mar 16th 2010, 02:52 PMtonio

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

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

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

Tonio - Mar 16th 2010, 02:53 PMMoo
Hello,

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

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

Your integral becomes:

$\displaystyle \int wcos(w) ~dw$

That is easily integrable by parts.

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

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

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

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

And there you go. - Mar 16th 2010, 03:21 PMblackwrx
Where did the $\displaystyle 2xcos(x)$ come from? What happened to the original $\displaystyle 2x^3$?

- Mar 16th 2010, 03:23 PMblackwrx
Nevermind, I see with the substitution that dw=2x times the w=x^2 does indeed solve my problem. Thanks guys.