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|dw:1442195407457:dw|

would it be like this so far?
|dw:1442195517625:dw| ? not too sure about the other parts :/

\(\large\rm t^2\quad\to\quad 2t\quad\to\quad 0\)

yes :)

Oh woops, lol not theta

\[\large\rm u=t^2\]\[\large\rm dv=\sin\beta t~d t\]

oops and now i did it!!
v = -cos BÃ¸ dt
lol

Hmm the \(\large\rm v\) looks a little off.

\[\large\rm \int\limits \cos(2x)dx=?\]Like this one, can you solve without u-sub?

:/ how would i fix v then?
but du = 2t ?

yes

du = 2t dt

ohh okay!! and not sure about about your above question :/

yes!!

ohh okay!! and so for our v, how can we apply the same method?

ohh okay!! and so we have that as our v?
so we do the integral udv = uv - integral v du ?

Yes, integral time \c;/

ooh okay!! and so we get this?|dw:1442196531964:dw|

-1/beta cos (beta t)t^2 -
not sure how to integrate that :/ but 2t dt = 0?

yes:)

and what happens next? we integrate the red?

okay!!

Any confusion there? :U
The problem is getting pretty long, so it's easy to get lost lol

So draw your by-parts thing again :D

|dw:1442197278644:dw|

okay :) let's see if i understood correctly lol :P
so we have this? |dw:1442197308540:dw|

|dw:1442197426918:dw|

okay! and so now i simplify ?

b=Beta, typo :c

I can type it out if you want lol
It's getting so long >.<

hehe okay!! thank you!! that would be very helpful :P I'm getting a bit lost in it lol :P

oooh wow okay!! haha got so messy :P so integrate it to -cos dt ?

Yes. And again, since we have a coefficient on the t, we have to divide by that.

how does that look again?

whoahhh wowzers okay :P
so this is done? :O

yes, imma check my work real quick though,
to make sure i didn't mess up anywhere

okie :)

ooh okay!! yay!! thanks so much!! :D