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roselin Group Title

elavualte indefinite integrals 1/x^(1/2) cos.( t^1/2 +3) dx

  • one year ago
  • one year ago

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  1. zepdrix Group Title
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    \[\huge \int\limits \frac{1}{x^{1/2}}\cdot \cos(t^{1/2}+3) \; dx\]Like this? Is that t suppose to be an x by chance?

    • one year ago
  2. roselin Group Title
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    yes, sorry. that is one t^1/2 not x^1/2

    • one year ago
  3. roselin Group Title
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    *that one is

    • one year ago
  4. zepdrix Group Title
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    \[\large \int\limits\limits \frac{1}{t^{1/2}}\cdot \cos(t^{1/2}+3) \; dt\] Oh ok :) Hmm looks like you can do a nice U-substitution for this one.

    • one year ago
  5. roselin Group Title
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    okay,

    • one year ago
  6. zepdrix Group Title
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    So with a U-sub, we're trying to find a U and U' somewhere in the problem. If you look at the inside of the cosine, taking the derivative of that should give you something similar to the other term. So try letting,\[\large u=t^{1/2}+3\]And see if you can match up your dU from there.

    • one year ago
  7. zepdrix Group Title
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    Were you able to get a dU? Or confused? :)

    • one year ago
  8. roselin Group Title
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    kind of confused

    • one year ago
  9. roselin Group Title
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    but trying to figure it out

    • one year ago
  10. zepdrix Group Title
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    |dw:1355686904344:dw|We're trying to do something like this, Simplify the integral so it's very easy to integrate. We just have to figure out what DU is so we can make sure we didn't miss anything.

    • one year ago
  11. zepdrix Group Title
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    If we let,\[\large u=t^{1/2}+3\]Taking the derivative with respect to t gives us,\[\large \frac{du}{dt}=\frac{1}{2t^{1/2}}\]

    • one year ago
  12. roselin Group Title
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    for du, we just eed to take the derivative of 1/t^1/2?

    • one year ago
  13. zepdrix Group Title
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    No, we want to take the derivative of the other part :O See what I did with u?

    • one year ago
  14. roselin Group Title
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    yes

    • one year ago
  15. zepdrix Group Title
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    From there, there is a process that allows us to write the dt on the other side. You can simply think of it as multiplication for your purposes. \[\large \frac{du}{dt}=\frac{1}{2t^{1/2}} \qquad \rightarrow \qquad \large du=\frac{1}{2t^{1/2}}dt\]

    • one year ago
  16. roselin Group Title
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    okay,

    • one year ago
  17. zepdrix Group Title
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    We're trying to replace this big chunk of stuff, Think of THIS as your variable that you're trying to solve for,\[\large \frac{1}{t^{1/2}}dt\] But we have a 1/2 in attached to it, how do we fix that?

    • one year ago
  18. roselin Group Title
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    can write it as square root of t?

    • one year ago
  19. roselin Group Title
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    not sure

    • one year ago
  20. zepdrix Group Title
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    oh i don't mean that 1/2. I mean the 2 in front of the t,\[\large \frac{1}{2t^{1/2}}\]See how this doesn't quite match what I posted before?

    • one year ago
  21. roselin Group Title
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    yeah

    • one year ago
  22. roselin Group Title
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    what do I do with it?

    • one year ago
  23. zepdrix Group Title
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    |dw:1355687571473:dw|

    • one year ago
  24. zepdrix Group Title
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    We'll multiply both sides by 2, to get rid of the 1/2.\[\large 2du=\frac{1}{t^{1/2}}dt\]

    • one year ago
  25. roselin Group Title
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    okay,

    • one year ago
  26. zepdrix Group Title
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    \[\large u=t^{1/2}+3 \qquad \qquad 2du=\frac{1}{t^{1/2}}dt\]Plugging these into our integral gives us,\[\large \int\limits \cos(t^{1/2}+3)\left(\frac{1}{t^{1/2}}dt\right) \qquad \rightarrow \qquad \int\limits \cos(u)(2du)\]

    • one year ago
  27. roselin Group Title
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    okay. i see

    • one year ago
  28. roselin Group Title
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    2 sin(t^1/2 + 3)?

    • one year ago
  29. zepdrix Group Title
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    Yesss very good.

    • one year ago
  30. roselin Group Title
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    |dw:1355688113175:dw|

    • one year ago
  31. roselin Group Title
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    in the second step , do I have to write the du?

    • one year ago
  32. zepdrix Group Title
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    If you still have the Integral bar, then yes. When the integral bar disappears, the du disappears also. It's a part of the integration process.

    • one year ago
  33. roselin Group Title
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    oh okay

    • one year ago
  34. roselin Group Title
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    i have another problem, which I am trying to solve, i will do my best to try to solve it, will you guide me through that, i will only ask you to check whether it is correct or not.

    • one year ago
  35. zepdrix Group Title
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    sure c: Imma go make some lunch real quick. So i brb. You can post some stuff though, and I'll take a look in a few minutes when i get back :D

    • one year ago
  36. roselin Group Title
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    sure, i need to have lunch too. ...

    • one year ago
  37. zepdrix Group Title
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    hah XD

    • one year ago
  38. roselin Group Title
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    |dw:1355688628735:dw|

    • one year ago
  39. roselin Group Title
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    i was trying to simplify it. is it correct?

    • one year ago
  40. zepdrix Group Title
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    yah looks good so far c:

    • one year ago
  41. zepdrix Group Title
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    Hmm I don't think this one will work out nicely with a U-sub. Have you learned about Trig-sbustitutions yet?

    • one year ago
  42. roselin Group Title
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    |dw:1355689271796:dw|

    • one year ago
  43. roselin Group Title
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    no, not yet

    • one year ago
  44. zepdrix Group Title
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    Hmm

    • one year ago
  45. roselin Group Title
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    i always get confused when calculating the du

    • one year ago
  46. zepdrix Group Title
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    \[\large u=x^2-1 \qquad \rightarrow \qquad du=2x \; dx\]It's just the derivative! :) Hmm I don't think this is going to help us though :( Hmmmm Thinking :3

    • one year ago
  47. roselin Group Title
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    so the one du that i have calculated is correct for the above problem?

    • one year ago
  48. zepdrix Group Title
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    No, I'm not really sure what you did there. You picked something for your U. Then instead of taking the derivative of U, you took the derivative of x^4.

    • one year ago
  49. roselin Group Title
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    ohh...

    • one year ago
  50. roselin Group Title
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    u= x^ 2 -1 , then du = 2x

    • one year ago
  51. roselin Group Title
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    sorry

    • one year ago
  52. roselin Group Title
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    2xdx

    • one year ago
  53. zepdrix Group Title
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    yah good times :) Unfortunately, I don't think this is going to help us :( We don't have any xdx in our problem, we have 1/x^4 dx but that's not the same thing. I think you NEED a trig sub for this one.. I could be wrong though D: This is a homework problem..?

    • one year ago
  54. roselin Group Title
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    yeah

    • one year ago
  55. roselin Group Title
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    can i use the wolframalpha and see the answer ? :)

    • one year ago
  56. roselin Group Title
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    i know i will get the answer there, but I need to understand how to do it.

    • one year ago
  57. zepdrix Group Title
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    http://www.wolframalpha.com/input/?i=integrate+%28x%5E2-1%29%5E%281%2F2%29%2Fx%5E4+dx Yah if you click the "show steps" button, you'll see that wolfram uses a trig sub to solve it. There might be another way, like by parts or something.. But I dunno :\

    • one year ago
  58. zepdrix Group Title
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    "step-by-step solution" button *

    • one year ago
  59. roselin Group Title
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    okay,

    • one year ago
  60. roselin Group Title
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    we did not learn the trig substitution yet and if I do that problem using the trig substitution I will get a 0 :)

    • one year ago
  61. roselin Group Title
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    the solution is right in front of me step by step .hmm

    • one year ago
  62. roselin Group Title
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    i will go by that rule only,

    • one year ago
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