Evaluate the integral inside! thanks!!

- anonymous

Evaluate the integral inside! thanks!!

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- schrodinger

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- anonymous

|dw:1442195407457:dw|

- anonymous

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

- zepdrix

This is how I like to think of it when doing `by-parts`.
\(\large\rm u\) 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.

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## More answers

- zepdrix

So 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

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

- anonymous

yes :)

- zepdrix

\[\large\rm u=t^2\]\[\large\rm dv=\sin\beta\theta~d\theta\]So what do we get for these \[\large\rm du=?\]\[\large\rm v=?\]

- zepdrix

Oh woops, lol not theta

- zepdrix

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

- anonymous

hahaa :P
du = 2t and v = -cosBtdt
^sorry!! using a B instead because i cannot find how to put the symbol in

- anonymous

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

- zepdrix

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

- zepdrix

You `could` do a little u-sub, err we already used the letter u....
you could do a little m-sub to deal with integrating your dv.
but that's really tedious, you wanna get comfortable with easy integrals like this.

- zepdrix

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

- anonymous

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

- zepdrix

yes

- zepdrix

du = 2t dt

- anonymous

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

- zepdrix

You 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?

- anonymous

yes!!

- zepdrix

Take 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

So 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

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

- zepdrix

\[\large\rm \int\limits \sin\beta t~d t=\frac{1}{\beta}(-\cos\beta t)\]Going backwards, sine gives us -cosine
and we have to divide by the beta coefficient on x.

- anonymous

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

- zepdrix

\[\large\rm u=t^2\qquad\qquad\qquad\qquad du=2t~d t\]\[\large\rm dv=\sin(\beta t)d t\qquad\qquad v=-\frac{1}{\beta}\cos(\beta t)\]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

Yes, integral time \c;/

- anonymous

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

- anonymous

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

- zepdrix

Ok good :)
Looks like `integration-by-parts` 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

And then we need to apply `by-parts` 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

yes:)

- anonymous

and what happens next? we integrate the red?

- zepdrix

The 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

I 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

Getting your other two pieces give you,\[\large\rm du=1dt\]\[\large\rm v=\frac{1}{\beta}\sin(\beta t)\]

- anonymous

okay!!

- zepdrix

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

- zepdrix

So draw your by-parts thing again :D

- zepdrix

|dw:1442197278644:dw|

- anonymous

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

- zepdrix

|dw:1442197426918:dw|

- anonymous

okay! and so now i simplify ?

- zepdrix

\[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\color{orangered}{\int\limits\limits\limits t \cos(\beta t) d t}\]
\[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\color{orangered}{\left[\frac{1}{\beta}t \sin(\beta t)-\frac{1}{b}\int\limits \sin(\beta t)d t\right]}\]Yes, maybe distribute the 2/B.

- zepdrix

b=Beta, typo :c

- zepdrix

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

- anonymous

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

- zepdrix

\[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)-\frac{2}{\beta^2}\int\limits \sin(\beta t)d t\]Something like that.
Ok one last piece to integrate! No more `by-parts` required, yay!

- anonymous

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

- zepdrix

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

- anonymous

how does that look again?

- zepdrix

\[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)-\frac{2}{\beta^2}\color{orangered}{\int\limits\limits \sin(\beta t)d t}\]
\[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)-\frac{2}{\beta^2}\color{orangered}{\frac{1}{\beta}(-\cos(\beta t))}\]

- zepdrix

\[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)+\frac{2}{\beta^3}\cos(\beta t)+C\]Somethinggggg like that? 0_o
Boy this one is a doozy when they throw in the beta and the squared term.

- anonymous

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

- zepdrix

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

- anonymous

okie :)

- zepdrix

yayyyy 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

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

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