# Thread: Integration by parts, HELP!

1. ## Integration by parts, HELP!

I'm supposed to integrate this:

(x^5)cos(x^3)

I understand to make u = x^5 in order to try and eliminate it, but that forces me to make dv = cos(x^3), which I am at a complete loss as to its integral. I snooped around on the net, and from some of my readings, it seems as though it requires a math program to do so(atleast the definite integral). My Ti-89 is of no help, so I come here...

2. Originally Posted by phack
I'm supposed to integrate this:

(x^5)cos(x^3)
This is a hard integral.
If you call $t=x^3$ then $t'=3x^2$ thus you need an $x^2$.

You split is as,
$\frac{1}{3} \int x^3 (3x^2) \cos x^3 dx$
Use substitution rule,
$\frac{1}{3} \int t\cos t dt$
Now, you need to do integration by parts.
Let $u=t$ and $v'=\cos t$.
Thus, $u'=1$ and $v=\sin t$.
Integration by parts,
$t\sin t - \int \sin t dt=t\sin t+\cos t+C$
Substitute back (and multiply by 1/3 from beginning),
$\frac{1}{3}x^3 \sin x^3+\frac{1}{3}\cos x^3 +C$
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If you want a website to check your integration work,
You can use The Integrator--Integrals from Mathematica

3. Thankyou, you are a life saver...