# Math Help - Integration by parts

1. ## Integration by parts

$\displaystyle\int(w^2-6w)sin(2w)dw$

u= $(w^2-6w)$ du= $(2w-6) dw$

dv= $sin(2w)dw$ and v= $\frac{-1}{2}cos(2w)$

$\displaystyle\frac{-1}{2}(w^2-6w)sin(2w)-\int \frac{-1}{2}(2w-6)cos(2w)dw$

u= $(2w-6)$ du= $2 dw$

dv= $cos(2w)dw$ and v= $\frac{1}{2}sin(2w)$

$\displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)-\int\frac{1}{2}sin(2w)*2dw$

$\displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)-\int\sin(2w)dw$

$\displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)+\frac{1}{2}cos(2w)+c$

Is my algebra wrong? Cause the book is saying that the answer is:

$\displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)+\frac{1}{4}cos(2w)+c$

2. Originally Posted by dbakeg00
$\displaystyle\int(w^2-6w)sin(2w)dw$

u= $(w^2-6w)$ du= $(2w-6) dw$

dv= $sin(2w)dw$ and v= $\frac{-1}{2}cos(2w)$

$\displaystyle\frac{-1}{2}(w^2-6w)sin(2w)-\int \frac{-1}{2}(2w-6)cos(2w)dw$
$v=\frac{-1}{2}cos(2w) \;and\; u= (w^2-6w) \implies uv\;=\;\frac{-1}{2}cos(2w)\times (w^2-6w)$

$\displaystyle \int u\;dv = uv - \intv\;du = \dfrac{-(w^2-6w)\cos(2w)}{2} + \int \frac{1}{2}cos(2w)(2w-6) dw$

3. I'm sorry, I don't understand what you just did there. I thought the formula for the integration by parts is $\displaystyle\int u\ dv=uv-\int v\ du$

4. Originally Posted by dbakeg00
I'm sorry, I don't understand what you just did there. I thought the formula for the integration by parts is $\displaystyle\int u\ dv=uv-\int v\ du$
It is..

So look at your integration by parts.. You have written them down on the first and second lines of your answer..

So what is uv??

instead of writing $uv= \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)$, you have written $uv=\displaystyle\frac{-1}{2}(w^2-6w)sin(2w)$

5. Originally Posted by dbakeg00
$\displaystyle\frac{-1}{2}(w^2-6w)sin(2w)-\int \frac{-1}{2}(2w-6)cos(2w)dw$

u= $(2w-6)$ du= $2 dw$

dv= $cos(2w)dw$ and v= $\frac{1}{2}sin(2w)$

$\displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)-\int\frac{1}{2}sin(2w)*2dw$
You are not handling the $-\frac{1}{2}$ term properly in the integral in the top line. Since you are not including this when integrating by parts, I assume you are treating it like this:

$\displaystyle-\frac{1}{2}(w^2-6w)\sin(2w)+\frac{1}{2} \left[ \int (2w-6)\cos(2w) \, dw \right]$

After integrating by parts, you are making a substitution in place of the integral in the above line (the part in brackets). So, after your substitution, you get this:

$\displaystyle-\frac{1}{2}(w^2-6w)\sin(2w)+\frac{1}{2} \left[ \frac{1}{2}(2w-6)\sin(2w) - \int \sin(2w) \, dw \right]$

Does this help?

6. Originally Posted by harish21
It is..

So look at your integration by parts.. You have written them down on the first and second lines of your answer..

So what is uv??

instead of writing $uv= \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)$, you have written $uv=\displaystyle\frac{-1}{2}(w^2-6w)sin(2w)$
What you are identifying was just a typo. Notice that he didn't carry that error through the remainder of his work.

7. Ahh..I see it now. Thanks for helping me out guys...I really appreciate it!