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roselin

  • 2 years ago

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

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  1. zepdrix
    • 2 years ago
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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?

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

  3. roselin
    • 2 years ago
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    *that one is

  4. zepdrix
    • 2 years ago
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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.

  5. roselin
    • 2 years ago
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    okay,

  6. zepdrix
    • 2 years ago
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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.

  7. zepdrix
    • 2 years ago
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    Were you able to get a dU? Or confused? :)

  8. roselin
    • 2 years ago
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    kind of confused

  9. roselin
    • 2 years ago
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    but trying to figure it out

  10. zepdrix
    • 2 years ago
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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.

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

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

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

  14. roselin
    • 2 years ago
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    yes

  15. zepdrix
    • 2 years ago
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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. \[\large \frac{du}{dt}=\frac{1}{2t^{1/2}} \qquad \rightarrow \qquad \large du=\frac{1}{2t^{1/2}}dt\]

  16. roselin
    • 2 years ago
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    okay,

  17. zepdrix
    • 2 years ago
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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?

  18. roselin
    • 2 years ago
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    can write it as square root of t?

  19. roselin
    • 2 years ago
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    not sure

  20. zepdrix
    • 2 years ago
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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?

  21. roselin
    • 2 years ago
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    yeah

  22. roselin
    • 2 years ago
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    what do I do with it?

  23. zepdrix
    • 2 years ago
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    |dw:1355687571473:dw|

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

  25. roselin
    • 2 years ago
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    okay,

  26. zepdrix
    • 2 years ago
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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,\[\large \int\limits \cos(t^{1/2}+3)\left(\frac{1}{t^{1/2}}dt\right) \qquad \rightarrow \qquad \int\limits \cos(u)(2du)\]

  27. roselin
    • 2 years ago
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    okay. i see

  28. roselin
    • 2 years ago
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    2 sin(t^1/2 + 3)?

  29. zepdrix
    • 2 years ago
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    Yesss very good.

  30. roselin
    • 2 years ago
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    |dw:1355688113175:dw|

  31. roselin
    • 2 years ago
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    in the second step , do I have to write the du?

  32. zepdrix
    • 2 years ago
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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.

  33. roselin
    • 2 years ago
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    oh okay

  34. roselin
    • 2 years ago
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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.

  35. zepdrix
    • 2 years ago
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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

  36. roselin
    • 2 years ago
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    sure, i need to have lunch too. ...

  37. zepdrix
    • 2 years ago
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    hah XD

  38. roselin
    • 2 years ago
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    |dw:1355688628735:dw|

  39. roselin
    • 2 years ago
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    i was trying to simplify it. is it correct?

  40. zepdrix
    • 2 years ago
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    yah looks good so far c:

  41. zepdrix
    • 2 years ago
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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?

  42. roselin
    • 2 years ago
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    |dw:1355689271796:dw|

  43. roselin
    • 2 years ago
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    no, not yet

  44. zepdrix
    • 2 years ago
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    Hmm

  45. roselin
    • 2 years ago
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    i always get confused when calculating the du

  46. zepdrix
    • 2 years ago
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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

  47. roselin
    • 2 years ago
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    so the one du that i have calculated is correct for the above problem?

  48. zepdrix
    • 2 years ago
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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.

  49. roselin
    • 2 years ago
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    ohh...

  50. roselin
    • 2 years ago
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    u= x^ 2 -1 , then du = 2x

  51. roselin
    • 2 years ago
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    sorry

  52. roselin
    • 2 years ago
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    2xdx

  53. zepdrix
    • 2 years ago
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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..?

  54. roselin
    • 2 years ago
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    yeah

  55. roselin
    • 2 years ago
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    can i use the wolframalpha and see the answer ? :)

  56. roselin
    • 2 years ago
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    i know i will get the answer there, but I need to understand how to do it.

  57. zepdrix
    • 2 years ago
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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 :\

  58. zepdrix
    • 2 years ago
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    "step-by-step solution" button *

  59. roselin
    • 2 years ago
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    okay,

  60. roselin
    • 2 years ago
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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 :)

  61. roselin
    • 2 years ago
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    the solution is right in front of me step by step .hmm

  62. roselin
    • 2 years ago
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    i will go by that rule only,

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