We didn't cover this in class all year, so I always have trouble with it. I just don't really know how to attack it at all.

Ps. What happened here the last day or two? I tried to post, kept getting 'Bandwidth Exceeded'.

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- Oct 29th 2008, 03:42 PMNaurIntegration by recognition
We didn't cover this in class all year, so I always have trouble with it. I just don't really know how to attack it at all.

Ps. What happened here the last day or two? I tried to post, kept getting 'Bandwidth Exceeded'. - Oct 29th 2008, 04:14 PMMathstud28
So we are given $\displaystyle f(x)=x\cos(3x)$ and

$\displaystyle f'(x)=\cos(3x)-3x\sin(3x)$

Or

$\displaystyle x\sin(3x)=\frac{1}{3}\bigg[\cos(3x)-f'(x)\bigg]$

integrating both sides gives

$\displaystyle \int{x}\sin(3x)dx=\frac{1}{3}\int\bigg[\cos(3x)-f'(x)\bigg]dx$

$\displaystyle =\frac{1}{9}\sin(3x)-\frac{1}{3}f(x)$

$\displaystyle =\frac{1}{9}\sin(3x)-\frac{1}{3}x\cos(3x)$ - Oct 29th 2008, 06:49 PMNaur
Woah. So what you've done is rearrange the information we're given...to make the equation we're supposed to be integrating the subject, then integrated it...That's amazing! That's incredible!

Thanks very much. - Oct 30th 2008, 12:19 PMMathstud28
- Nov 1st 2008, 02:19 AMNaur
You're

*my*hero.

*looks admirably* - Nov 8th 2008, 04:01 PMNaur
Hold on. When you did the step 'integrating both sides', you didn't integrate the 1/3. Why is that?

- Nov 8th 2008, 04:32 PMChop Suey
- Nov 8th 2008, 04:52 PMNaur
Oh I see.

But if I had, like, 1 + Sin(3x) in that integral statement, and I was to take that out, it'd become antidifferentiated to x.

Thanks :D

Answers questions I didn't even know I had. I've done a few similar examples where I just took the logic the book used for granted, only now do I realize what I thought I knew was contradictory to itself. - Nov 8th 2008, 09:08 PMMathstud28
- Nov 8th 2008, 09:13 PMNaur
I'd never seen that before, but on here it's not uncommon to see unfamiliar notation. So I assumed that's what it meant. :)