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
$\displaystyle \int{x}\sin(3x)dx=\frac{1}{3}\int\bigg[\cos(3x)-f'(x)\bigg]dx$
$\displaystyle =\frac{1}{9}\sin(3x)-\frac{1}{3}f(x)$
$\displaystyle =\frac{1}{9}\sin(3x)-\frac{1}{3}x\cos(3x)$
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.