Find the integral.
im sure u = x^5 and v'= cos(x^3) but not sure where to go from there
well, from there you would do the integration by parts method, but as it stands, you'd have to do it 5 times!
a preliminary substitution makes the problem easier:
Let $\displaystyle t = x^3$
$\displaystyle \Rightarrow dt = 3x^2 ~dx$
$\displaystyle \Rightarrow \frac 13 ~dt = x^2~dx$
So our integral becomes
$\displaystyle \frac 13 \int t \cos t~dt$
which is a relatively easy integration by parts problem
Hello, gabet16941!
Break it up like this:$\displaystyle \int x^5\cos(x^3)\,dx$
I'm sure $\displaystyle u = x^5\,\text{ and }\,v' = \cos(x^3)$ . . . . but you can't integrate $\displaystyle {\color{blue}\cos(x^3)}$
. . $\displaystyle \begin{array}{ccccccc}u &=& x^3 & & dv &=& x^2\cos(x^3)\,dx \\ du &=& 3x^2\,dx & & v &=& \frac{1}{3}\sin(x^3) \end{array}$
Then we have: .$\displaystyle \tfrac{1}{3}x^3\sin(x^3) - \tfrac{1}{3}\int x^2\sin(x^3)\,dx $
. . . . . . . . . $\displaystyle = \;\tfrac{1}{3}x^3\sin(x^3) + \tfrac{1}{9}\cos(x^3) + C$