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anonymous

  • 5 years ago

Integral of 1/(sqrt(9-4x^2))

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    use wolfram alpha

  3. anonymous
    • 5 years ago
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    It doesn't show any of the steps.

  4. anonymous
    • 5 years ago
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    it does, click show steps

  5. nowhereman
    • 5 years ago
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    Substitute \(\frac{2}{3}x = \sin z\)

  6. anonymous
    • 5 years ago
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    nowhereman, I got that part. My only problem is getting the 1/2 in front of the integral.

  7. anonymous
    • 5 years ago
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    nowhereman, revolve the region bounded by y = (x+2)^2 , y = 2, y=0, about the y axis (using shell method). i get a negative v0lume

  8. anonymous
    • 5 years ago
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    laplace... http://i301.photobucket.com/albums/nn63/knightfire8/Screenshot2011-05-26at104452PM.png It doesn't show the steps first of all. Second..Im trying to ask a simple question, please dont use my question as your own space.

  9. anonymous
    • 5 years ago
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    yes it does show the steps

  10. anonymous
    • 5 years ago
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    its one step

  11. anonymous
    • 5 years ago
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    It's definitely not one step. I have at least 4 to get to my answer which has everything except the 1/2. Now please move on.

  12. anonymous
    • 5 years ago
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    its one ugly giant step

  13. anonymous
    • 5 years ago
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    then pay for a tutor, , and dont tell me what to do

  14. anonymous
    • 5 years ago
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    check pauls notes on trig substitution, whatever

  15. nowhereman
    • 5 years ago
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    laplace, could you please post your question seperately, I will help knightfire

  16. nowhereman
    • 5 years ago
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    Could you show the substitution step you are doing knightfire, that's where the 1/2 comes from.

  17. anonymous
    • 5 years ago
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    So I set x = 3/2(sinz), dx = 3coszdz, and sqrt(9-4x^2) = 3cosz

  18. anonymous
    • 5 years ago
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    I plug them in for their appropriate parts and I get z +c

  19. nowhereman
    • 5 years ago
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    aha, that's the mistake, dx = 3/2 cos z dz

  20. anonymous
    • 5 years ago
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    Wait, so how did you get the 1/2 you multiple by? That's the part Im confused on. Because as mentioned before I got everything the same as the answer it listed except for 1/2.

  21. anonymous
    • 5 years ago
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    multiply*

  22. nowhereman
    • 5 years ago
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    Because if you differentiate \(x\) you will get \(\mathrm{d}(\frac{3}{2}\sin z)/\mathrm{dz} = \frac{3}{2} \cos z\) and that's where your \(1/2\) is missing.

  23. anonymous
    • 5 years ago
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    Ohhh ok that makes sense now. I guess I forgot that x and dx were directly related to each other with all these trig equations going on...Thanks so much nowhereman, you were most helpful.

  24. anonymous
    • 5 years ago
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    ok

  25. nowhereman
    • 5 years ago
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    laplace, you might get a negative volume, because the area you are rotation is on the left side of the y-axis. Try to mirror it, i.e. you \(y \leq (x-2)^2\) and the interval \([0, 2]\)

  26. anonymous
    • 5 years ago
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    \[\int\limits{1/\sqrt{(9-4x ^{2})}}dx\] \[=1/2 \int\limits 1/\sqrt{((3/2)^{2}-x ^{2})} dx\] \[= (1/2)*2/(2*3) \ln ((3/2+x)/(3/2-x)) +c\] \[=1/6 \ln ((3+2x)/(3-2x))+c\]

  27. anonymous
    • 5 years ago
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    \[\int\limits \frac{1}{\sqrt{9 - 4*x^2}} \, dx=\frac{1}{2} \text{ArcSin}\left[\frac{2 x}{3}\right]+c \]

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