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A very tuf integration problem: Urgent Help Required

  1. Nov 29, 2005 #1
    Hi guys!

    well im stuck at the following problem and donot know how to proceed.

    Plz help me

    here is the problem

    [tex]\int\sqrt{\tan{\theta}}d\theta[/tex]

    here is what im doing

    let

    [tex]u=\tan{\theta}[/tex]

    [tex]du=(\sec^2{\theta}) d\theta[/tex]

    [tex]1+u^2=(\sec^2{\theta})[/tex]

    therefore

    [tex]du= (1+u^2)d\theta[/tex]


    so now i can write:

    [tex]\int\frac{\sqrt{u}}{1+u^2}du[/tex]


    but now im stuck that how should i solve this problem now

    am i applying the right technique then what should i do now and if im wrong then what is the right method.

    PLz help me as soon as possible as i have just hours left now.

    Thanks in advance
     
    Last edited: Nov 29, 2005
  2. jcsd
  3. Nov 29, 2005 #2

    TD

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    Now substituting [itex]v = \sqrt{u}[/itex], or directly doing [itex]v = \sqrt{\tan{\theta}}[/itex], would give:

    [tex]\int \frac{2v^2}{1+v^4} dv[/tex]

    Which gets rid of the square root but of course, introduces higher powers.
     
  4. Nov 29, 2005 #3

    siddharth

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    Instead of [tex] u=\tan\theta [/tex]
    Try

    [tex] u^2=\tan\theta [/tex]

    Once you make that substitution into the integral, divide the numerator and denominator by [tex] u^2 [/tex].

    Then your denominator will be of the form [tex] u^2 + \frac{1}{u^2} [/tex].

    Complete the square and see if you can manipulate the numerator (by adding and subtracting). It should be obvious once you do that.
     
    Last edited: Nov 29, 2005
  5. Nov 29, 2005 #4

    GCT

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    yeah, just do what TD suggested and then you've got some nasty partial fraction work to do, tis one way to solve it....
     
  6. Nov 29, 2005 #5
    Also, there were a few other threads on this a while back you might want to check out if you have any more questions.
     
  7. Nov 30, 2005 #6

    siddharth

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    GCT, you can avoid partial fractions. All you have to do is add and subtract [tex] \frac{1}{u^2} [/tex] to the numerator and write the denominator as [tex] (u+\frac{1}{u})^2 -2 [/tex] and [tex] (u-\frac{1}{u})^2 + 2[/tex].
     
  8. Dec 2, 2005 #7

    GCT

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    At this moment I'm not quite sure what you're referring to, I apologize but right now I don't have time to work out your proposal. If you really wish to make your case evident, you should write down the full steps towards the full solution....you'll get more comments that way.
     
  9. Dec 3, 2005 #8

    siddharth

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    From

    [tex] \int \frac{2u^2}{1+u^4} du [/tex]

    [tex] \int \frac{2}{u^2+\frac{1}{u^2}} du [/tex]

    [tex] \int (\frac{1+\frac{1}{u^2}}{u^2 + \frac{1}{u^2}} + \frac{1-\frac{1}{u^2}}{u^2 + \frac{1}{u^2}}) du [/tex]


    [tex] \int (\frac{1+\frac{1}{u^2}}{(u-\frac{1}{u})^2 + 2} + \frac{1-\frac{1}{u^2}}{(u+\frac{1}{u})^2 - 2}) du [/tex]

    Evaluating each of these integrals is now very easy
     
    Last edited: Dec 3, 2005
  10. Dec 3, 2005 #9

    GCT

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    not VERY simple. Let's see your final solution. So far your integral is starting to resemble the partial fraction solution, that is the partial fraction solution process has the same simplifications and forms.
     
  11. Dec 3, 2005 #10

    siddharth

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    Why not?
    For the first Integral, set
    [tex] u - \frac{1}{u} = t [/tex]
    so that the Integral reduces to
    [tex] \int \frac{dt}{t^2+2} [/tex]
    For the second one, set
    [tex] u + \frac{1}{u} = t [/tex]
    and the Integral reduces to
    [tex] \int \frac{dt}{t^2 -2} [/tex]
    Surely this is easier?
     
  12. Dec 3, 2005 #11

    GCT

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    right, alright I get it....very nice.
     
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