I am pretty sure I need to use integration by parts, but I cannot get anywhere with this.
$\displaystyle \int 2x^3\cos(x^2)\,dx$
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
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.