Integration by recognition

**Naur** · October 29th 2008, 03:42 PM

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'.

**Mathstud28** · October 29th 2008, 04:14 PM

Originally Posted by Naur

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'.

So we are given $f(x)=x\cos(3x)$ and

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

Or

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

integrating both sides gives

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

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

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

**Naur** · October 29th 2008, 06:49 PM

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.

**Mathstud28** · October 30th 2008, 12:19 PM

Originally Posted by Naur

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.

::manly sniff of the nose:: its what I do

**Naur** · November 1st 2008, 02:19 AM

You're my hero.
*looks admirably*

**Naur** · November 8th 2008, 04:01 PM

Hold on. When you did the step 'integrating both sides', you didn't integrate the 1/3. Why is that?

**Chop Suey** · November 8th 2008, 04:32 PM

Originally Posted by Naur

Hold on. When you did the step 'integrating both sides', you didn't integrate the 1/3. Why is that?

$\int c f(x)~dx = c \int f(x)~dx = cF(x)$

For some constant c.

**Naur** · November 8th 2008, 04:52 PM

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

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.

**Mathstud28** · November 8th 2008, 09:08 PM

Originally Posted by Chop Suey

$\int c f(x)~dx = c \int f(x)~dx = cF(x)$

For some constant c.

This may sound stupid but don't forget to say $\text{where }F'(x)=f(x)$

**Naur** · November 8th 2008, 09:13 PM

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.

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