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anonymous
 one year ago
Evaluate the integral inside! thanks!!
anonymous
 one year ago
Evaluate the integral inside! thanks!!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442195407457:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it be like this so far? dw:1442195517625:dw ? not too sure about the other parts :/

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2This is how I like to think of it when doing `byparts`. \(\large\rm u\) is the thing that you're trying to "destroy" or breakdown. \(\large\rm dv\) is usually the part that is cyclical, won't breakdown, like sine/cosine or exponential.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So when I look at this problem, I'm thinking hmm \(\large\rm t^2\) is probably a good \(\large\rm u\) because we differentiate it and it will breakdown, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\rm t^2\quad\to\quad 2t\quad\to\quad 0\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm u=t^2\]\[\large\rm dv=\sin\beta\theta~d\theta\]So what do we get for these \[\large\rm du=?\]\[\large\rm v=?\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oh woops, lol not theta

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm u=t^2\]\[\large\rm dv=\sin\beta t~d t\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hahaa :P du = 2t and v = cosBtdt ^sorry!! using a B instead because i cannot find how to put the symbol in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops and now i did it!! v = cos Bø dt lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Hmm the \(\large\rm v\) looks a little off.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2You `could` do a little usub, err we already used the letter u.... you could do a little msub to deal with integrating your dv. but that's really tedious, you wanna get comfortable with easy integrals like this.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \int\limits \cos(2x)dx=?\]Like this one, can you solve without usub?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:/ how would i fix v then? but du = 2t ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay!! and not sure about about your above question :/

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2You can use a method which I think is called "advanced guessing",\[\large\rm \int\limits\limits \cos(2x)dx=\sin(2x)\]I'm thinking that it's going to be `something` like this, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Take a derivative to check and see if you have the correct solution.\[\large\rm (\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)\]Woops! We got a little bit too much back. See that extra 2? That tells us that:\[\large\rm \int\limits 2\cos(2x)dx=\sin(2x)\]Which means:\[\large\rm \int\limits \cos(2x)dx=\frac{1}{2}\sin(2x)\]Chain rule told us to multiply by 2 on the outside. So when we integrate, we need to do the reverse to compensate for the missing 2. We need to divide by 2.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So this will come up A LOT, so try to get used to it. When you have a coefficient on the x, you end up dividing by that coefficient when you integrate.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay!! and so for our v, how can we apply the same method?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \int\limits \sin\beta t~d t=\frac{1}{\beta}(\cos\beta t)\]Going backwards, sine gives us cosine and we have to divide by the beta coefficient on x.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay!! and so we have that as our v? so we do the integral udv = uv  integral v du ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm u=t^2\qquad\qquad\qquad\qquad du=2t~d t\]\[\large\rm dv=\sin(\beta t)d t\qquad\qquad v=\frac{1}{\beta}\cos(\beta t)\]Let's list all of our pieces together before setting up the integral. Unless you're following on paper, then you're prolly ok heh.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes, integral time \c;/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh okay!! and so we get this?dw:1442196531964:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01/beta cos (beta t)t^2  not sure how to integrate that :/ but 2t dt = 0?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Ok good :) Looks like `integrationbyparts` is required again. Clean things up before choosing your \(\large\rm u\) and \(\large\rm dv\) though:\[\large\rm =\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\int\limits t \cos(\beta t) d t\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2And then we need to apply `byparts` to this orange portion, ya?\[\large\rm =\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\color{orangered}{\int\limits\limits t \cos(\beta t) d t}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and what happens next? we integrate the red?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2The orange, yes. We need to establish \(\large\rm u\) and \(\large\rm dv\) for the orange part. Just a note: Earlier I had said something about how the t^2 will break down and I posted this:\[\large\rm t^2\quad\to\quad 2t\quad\to\quad 0\]Lil mistake there, 2t doesn't differentiate to 0, it differentiates to 2. Whatever, not a big deal :) We'll just take our derivative carefully

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I pulled the 2 outside of the integral, hopefully that isn't confusing. So we still want the u to break down. So we'll choose our parts this way:\[\large\rm u=t\]\[\large\rm dv=\cos(\beta t)d t\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Getting your other two pieces give you,\[\large\rm du=1dt\]\[\large\rm v=\frac{1}{\beta}\sin(\beta t)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Any confusion there? :U The problem is getting pretty long, so it's easy to get lost lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So draw your byparts thing again :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay :) let's see if i understood correctly lol :P so we have this? dw:1442197308540:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay! and so now i simplify ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm =\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\color{orangered}{\int\limits\limits\limits t \cos(\beta t) d t}\] \[\large\rm =\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\color{orangered}{\left[\frac{1}{\beta}t \sin(\beta t)\frac{1}{b}\int\limits \sin(\beta t)d t\right]}\]Yes, maybe distribute the 2/B.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I can type it out if you want lol It's getting so long >.<

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hehe okay!! thank you!! that would be very helpful :P I'm getting a bit lost in it lol :P

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm =\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)\frac{2}{\beta^2}\int\limits \sin(\beta t)d t\]Something like that. Ok one last piece to integrate! No more `byparts` required, yay!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oooh wow okay!! haha got so messy :P so integrate it to cos dt ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes. And again, since we have a coefficient on the t, we have to divide by that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how does that look again?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm =\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)\frac{2}{\beta^2}\color{orangered}{\int\limits\limits \sin(\beta t)d t}\] \[\large\rm =\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)\frac{2}{\beta^2}\color{orangered}{\frac{1}{\beta}(\cos(\beta t))}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm =\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)+\frac{2}{\beta^3}\cos(\beta t)+C\]Somethinggggg like that? 0_o Boy this one is a doozy when they throw in the beta and the squared term.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whoahhh wowzers okay :P so this is done? :O

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2yes, imma check my work real quick though, to make sure i didn't mess up anywhere

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2yayyyy looks like we did it correctly \c:/ I know that one was pretty insane to get through. You maybe would want to think about practicing with something like this:\[\large\rm \int\limits x^2\sin(2x)dx\]Might just be easier without all those betas floating around.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh okay!! yay!! thanks so much!! :D
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