Getting stuck on integration by parts. Here's a similar problem to the one I'm trying to work out:

Question

anonymous · Answer

$$\int\limits (x \cos(4x)) dx$$

anonymous · Answer

So if I understand correctly, I can look at this as two functions, where f(x)=cos(2x) and g'(x)=x.

anonymous · Answer

Ah, ok.

loser66 · Answer

The integration by part :
let u = x                                                               dv = cos (4x) dx
du = dx                                                                  v = $\dfrac {sin(4x)}{4}$
the part u is quite easy, right? the part v : to get v, you just integral dv , and that is the result. Got this part?

anonymous · Answer

Yes.

loser66 · Answer

then, just apply the formula

anonymous · Answer

now f(x)*g(x)-∫f'(x)g(x)?

loser66 · Answer

yup

anonymous · Answer

Almost there...(coding all day, my brain is a little sloshy. :)

loser66 · Answer

You can solve it, right? no stuck anymore, right?

anonymous · Answer

I thought so...

So I take {[sin(2x)/2]*x}-∫[sin(2x)/2]dx

loser66 · Answer

why 2x? the original problem is sin(4x) , right? why do you eat 2?

anonymous · Answer

Sorry, the original problem was 2x, I was trying to avoid the exact problem originally.

loser66 · Answer

hahaha... big OK

loser66 · Answer

next?

anonymous · Answer

I ended up with [(1/2)*x sin(x)]+[(1/4)cos(2x)]

loser66 · Answer

+C

anonymous · Answer

Gah! I think I need to take a break! ;) 2x

anonymous · Answer

+ C, yes. :)

anonymous · Answer

That's the most this secular guy has ever wanted to shout "bless you!" Thanks for your help, I really appreciate it.

loser66 · Answer

:)