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anonymous

  • one year ago

Evaluate the integral inside! thanks!!

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  1. anonymous
    • one year ago
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    |dw:1442195407457:dw|

  2. anonymous
    • one year ago
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    would it be like this so far? |dw:1442195517625:dw| ? not too sure about the other parts :/

  3. zepdrix
    • one year ago
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    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.

  4. zepdrix
    • one year ago
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    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?

  5. zepdrix
    • one year ago
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    \(\large\rm t^2\quad\to\quad 2t\quad\to\quad 0\)

  6. anonymous
    • one year ago
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    yes :)

  7. zepdrix
    • one year ago
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    \[\large\rm u=t^2\]\[\large\rm dv=\sin\beta\theta~d\theta\]So what do we get for these \[\large\rm du=?\]\[\large\rm v=?\]

  8. zepdrix
    • one year ago
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    Oh woops, lol not theta

  9. zepdrix
    • one year ago
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    \[\large\rm u=t^2\]\[\large\rm dv=\sin\beta t~d t\]

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

  11. anonymous
    • one year ago
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    oops and now i did it!! v = -cos Bø dt lol

  12. zepdrix
    • one year ago
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    Hmm the \(\large\rm v\) looks a little off.

  13. zepdrix
    • one year ago
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    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.

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

  15. anonymous
    • one year ago
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    :/ how would i fix v then? but du = 2t ?

  16. zepdrix
    • one year ago
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    yes

  17. zepdrix
    • one year ago
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    du = 2t dt

  18. anonymous
    • one year ago
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    ohh okay!! and not sure about about your above question :/

  19. zepdrix
    • one year ago
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    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?

  20. anonymous
    • one year ago
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    yes!!

  21. zepdrix
    • one year ago
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    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.

  22. zepdrix
    • one year ago
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    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.

  23. anonymous
    • one year ago
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    ohh okay!! and so for our v, how can we apply the same method?

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

  25. anonymous
    • one year ago
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    ohh okay!! and so we have that as our v? so we do the integral udv = uv - integral v du ?

  26. zepdrix
    • one year ago
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    \[\large\rm u=t^2\qquad\qquad\qquad\qquad du=2t~d t\]\[\large\rm dv=\sin(\beta t)d t\qquad\qquad v=-\frac{1}{\beta}\cos(\beta t)\]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.

  27. zepdrix
    • one year ago
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    Yes, integral time \c;/

  28. anonymous
    • one year ago
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    ooh okay!! and so we get this?|dw:1442196531964:dw|

  29. anonymous
    • one year ago
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    -1/beta cos (beta t)t^2 - not sure how to integrate that :/ but 2t dt = 0?

  30. zepdrix
    • one year ago
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    Ok good :) Looks like `integration-by-parts` is required again. Clean things up before choosing your \(\large\rm u\) and \(\large\rm dv\) though:\[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\int\limits t \cos(\beta t) d t\]

  31. zepdrix
    • one year ago
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    And then we need to apply `by-parts` to this orange portion, ya?\[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\color{orangered}{\int\limits\limits t \cos(\beta t) d t}\]

  32. anonymous
    • one year ago
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    yes:)

  33. anonymous
    • one year ago
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    and what happens next? we integrate the red?

  34. zepdrix
    • one year ago
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    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

  35. zepdrix
    • one year ago
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    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:\[\large\rm u=t\]\[\large\rm dv=\cos(\beta t)d t\]

  36. zepdrix
    • one year ago
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    Getting your other two pieces give you,\[\large\rm du=1dt\]\[\large\rm v=\frac{1}{\beta}\sin(\beta t)\]

  37. anonymous
    • one year ago
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    okay!!

  38. zepdrix
    • one year ago
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    Any confusion there? :U The problem is getting pretty long, so it's easy to get lost lol

  39. zepdrix
    • one year ago
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    So draw your by-parts thing again :D

  40. zepdrix
    • one year ago
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    |dw:1442197278644:dw|

  41. anonymous
    • one year ago
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    okay :) let's see if i understood correctly lol :P so we have this? |dw:1442197308540:dw|

  42. zepdrix
    • one year ago
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    |dw:1442197426918:dw|

  43. anonymous
    • one year ago
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    okay! and so now i simplify ?

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

  45. zepdrix
    • one year ago
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    b=Beta, typo :c

  46. zepdrix
    • one year ago
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    I can type it out if you want lol It's getting so long >.<

  47. anonymous
    • one year ago
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    hehe okay!! thank you!! that would be very helpful :P I'm getting a bit lost in it lol :P

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

  49. anonymous
    • one year ago
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    oooh wow okay!! haha got so messy :P so integrate it to -cos dt ?

  50. zepdrix
    • one year ago
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    Yes. And again, since we have a coefficient on the t, we have to divide by that.

  51. anonymous
    • one year ago
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    how does that look again?

  52. zepdrix
    • one year ago
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    \[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)-\frac{2}{\beta^2}\color{orangered}{\int\limits\limits \sin(\beta t)d t}\] \[\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)-\frac{2}{\beta^2}\color{orangered}{\frac{1}{\beta}(-\cos(\beta t))}\]

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

  54. anonymous
    • one year ago
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    whoahhh wowzers okay :P so this is done? :O

  55. zepdrix
    • one year ago
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    yes, imma check my work real quick though, to make sure i didn't mess up anywhere

  56. anonymous
    • one year ago
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    okie :)

  57. zepdrix
    • one year ago
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    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.

  58. anonymous
    • one year ago
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    ooh okay!! yay!! thanks so much!! :D

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