1. ## 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'.

2. 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

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

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

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

3. 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.

4. 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

5. You're my hero.

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

7. 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.

8. 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.

9. 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 $\displaystyle \text{where }F'(x)=f(x)$

10. 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.