# Thread: integration by parts question 2

1. ## integration by parts question 2

Find the integral.

im sure u = x^5 and v'= cos(x^3) but not sure where to go from there

2. Originally Posted by gabet16941
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 $t = x^3$

$\Rightarrow dt = 3x^2 ~dx$

$\Rightarrow \frac 13 ~dt = x^2~dx$

So our integral becomes

$\frac 13 \int t \cos t~dt$

which is a relatively easy integration by parts problem

3. Hello, gabet16941!

$\int x^5\cos(x^3)\,dx$

I'm sure $u = x^5\,\text{ and }\,v' = \cos(x^3)$ . . . . but you can't integrate ${\color{blue}\cos(x^3)}$
Break it up like this:

. . $\begin{array}{ccccccc}u &=& x^3 & & dv &=& x^2\cos(x^3)\,dx \\ du &=& 3x^2\,dx & & v &=& \frac{1}{3}\sin(x^3) \end{array}$

Then we have: . $\tfrac{1}{3}x^3\sin(x^3) - \tfrac{1}{3}\int x^2\sin(x^3)\,dx$

. . . . . . . . . $= \;\tfrac{1}{3}x^3\sin(x^3) + \tfrac{1}{9}\cos(x^3) + C$