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Integration of Sqrt Tangent

  1. Jun 18, 2005 #1
    How would you compute something like:

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

    I cannot seem to be able to use any trig. identities.

    Thanks.
     
  2. jcsd
  3. Jun 18, 2005 #2
    Try [tex]u = tan\theta [/tex]

    Then [tex]u = x^2[/tex]

    One more hint: [tex](x^4+1)= (x^2 - \sqrt{2}x+1)(x^2 + \sqrt{2}x + 1)[/tex]
     
  4. Jun 18, 2005 #3
    There was a very long thread about this integral already, click here
     
  5. Jun 18, 2005 #4
    Yeah, but then the new integral has two different variables.

    With [tex]u=\tan{\theta}[/tex], I come up with:

    [tex]\int \sqrt{u}\cos^2{\theta}du[/tex]
     
    Last edited: Jun 18, 2005
  6. Jun 18, 2005 #5

    dextercioby

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

    is the right one to do.

    Daniel.
     
  7. Jun 18, 2005 #6
    cos^2 = 1 / sec^2 right, and sec^2 = 1 + tan^2 right

    so the integral is

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

    then used the hint above.

    edit... looking at the other thread, you should probably do dex's substitution.
     
    Last edited: Jun 18, 2005
  8. Jun 18, 2005 #7
    Alright, that makes sense. The only problem I see is this:

    Shouldn't the integral be [tex]\int \frac{\sqrt{u}}{1+u^2}du[/tex]?
     
  9. Jun 18, 2005 #8
    Yes you are right, sorry.
     
  10. Jun 18, 2005 #9
    Alright, well I can integrate that. Thanks a lot for your help.
     
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