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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


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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.

    Try it out. Input Sqrt[Tan[x]]. Looks rather complicated :tongue2: .
  5. Jul 29, 2004 #4


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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


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    parth dave's substitution seems best to me.
    We have:

    We note the identity:

    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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