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Integral of (tan x)^(1/2)

  1. Sep 20, 2007 #1
    Can please anyone show me how to solve the

    integral of (tan x)^(1/2) ? (integral of the square root of tangent)

    Thank you in advance,
    Panos


    I can't do anything else than the obvious tan=sin/cos. I really can't figure out which differentiation is giving you the sin^(1/2) * cos^(-1/2) :(
     
    Last edited: Sep 21, 2007
  2. jcsd
  3. Sep 20, 2007 #2

    Dick

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    Start with substituting u^2=tan(x) and see where that leads you.
     
  4. Sep 21, 2007 #3

    Gib Z

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    Don't give up if the work gets a little complex btw. Personally I tend to think it's a sad sign when a simple looking integral gets really ugly, but this one gets REALLY ugly.
     
  5. Sep 21, 2007 #4
    It doesnt get ugly. Just keep at it. It just a bit long is all.
     
  6. Sep 21, 2007 #5

    CompuChip

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    And you probably need a little identity about cos(arctan(z)).
     
  7. Sep 21, 2007 #6

    Gib Z

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  8. Sep 22, 2007 #7
    Thank u all very much. :)
    With ur suggestions I got it until
    [tex]2\int(\frac{u^2}{1+u^4}du[/tex]

    but then, stuck :p

    I'm goin to see the solution now, thnx Gib for the link
     
  9. Dec 7, 2009 #8
    I got
    [tex]2\int\frac{u^2}{1+u^4}du = 2\int\frac{u^2}{\left(u^2 + \sqrt{2}u + 1\right)\left(u^2 - \sqrt{2}u + 1\right)}du[/tex]


    the partial fraction expansion doesn't look good. :(
     
  10. Dec 7, 2009 #9

    Redbelly98

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    Mod's note:
    Misleading nonsense has been deleted. PF regrets the confusion this may have caused.

    murshid_islam, I'll suggest starting a new thread as your best chance of getting help, since this thread was from 2 years ago. Again, we regret the misleading, confusing posts that were made here.
     
  11. Dec 10, 2009 #10
    Thanks.


    Thanks again for the advice. I will start a new thread.
     
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