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Calculus problem

  1. Jul 28, 2004 #1
    anyone has an easy way to do the following problem..
    [tex] \int \sqrt{\tan x} dx [/tex]
     
  2. jcsd
  3. Jul 29, 2004 #2

    mathwonk

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    the basic idea of substitution is to try to put a new symbol in for anything you find confusing. Like here you might try u = tan(x), and then of course you have to put du
    = sec^2(x)dx. See if that makes it any better. you may to do some more substitutions later.
     
  4. Jul 29, 2004 #3
    I don't think a substitution is going to do it.

    http://integrals.wolfram.com/index.en.cgi
    Try it out. Input Sqrt[Tan[x]]. Looks rather complicated :tongue2: .
     
  5. Jul 29, 2004 #4

    Zurtex

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    If [itex]du = \sec^2x dx[/itex] then:

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

    Hope that helps. Your integral is now:

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

    I'd look at using byparts from there on, but that's just a guess.
     
    Last edited: Jul 29, 2004
  6. Jul 29, 2004 #5

    arildno

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    parth dave's substitution seems best to me.
    We have:
    [tex]u=\sqrt{tan{x}}\to{du}=\frac{dx}{2u}\frac{1}{\cos^{2}x}=\frac{dx}{2u}(u^{4}+1)[/tex]
    Or:
    [tex]\int\sqrt{tan{x}}dx=\int\frac{2u^{2}}{u^{4}+1}du[/tex]

    We note the identity:
    [tex]u^{4}+1=(u^{2}-\sqrt{2}u+1)(u^{2}+\sqrt{2}u+1)[/tex]

    We may now use partial fractions techniques to derive the answer.
     
  7. Aug 7, 2004 #6
    Put tanx=t square

    then dx/dt=sec(square)t

    put sec(sqr)t= 1+t(sqr)

    and use partial fractions ...thats it..
     
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