# Integration by recognition

• Oct 29th 2008, 03:42 PM
Naur
Integration 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 PM
Mathstud28
Quote:

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 $\displaystyle f(x)=x\cos(3x)$ 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 PM
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.
• Oct 30th 2008, 12:19 PM
Mathstud28
Quote:

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
• Nov 1st 2008, 02:19 AM
Naur
You're my hero.
• Nov 8th 2008, 04:01 PM
Naur
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 PM
Chop Suey
Quote:

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

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

For some constant c.
• Nov 8th 2008, 04:52 PM
Naur
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 PM
Mathstud28
Quote:

Originally Posted by Chop Suey
$\displaystyle \int c f(x)~dx = c \int f(x)~dx = cF(x)$

For some constant c.

This may sound stupid but don't forget to say $\displaystyle \text{where }F'(x)=f(x)$
• Nov 8th 2008, 09:13 PM
Naur
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. :)