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Integral of sqrt(tan x)

  1. Dec 10, 2009 #1
    My question is about this integral:

    [tex]\int\sqrt{\tan (x)}dx[/tex]

    After using the substitution, u2 = tan(x), I got,

    [tex]2\int\frac{u^2}{u^4 + 1}du = 2\int\frac{u^2}{\left(u^2 + \sqrt{2}u + 1\right)\left(u^2 - \sqrt{2}u + 1\right)}du[/tex]

    Next, I tried the partial fraction expansion. But it turned pretty ugly. Is there any easier way of doing it?
     
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  3. Dec 10, 2009 #2

    nicksauce

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    If it is a definite integral, I think contour integration would be the way to go... otherwise I can't think of a way to make it nice.
     
  4. Dec 10, 2009 #3

    dextercioby

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    I don't think there's a neater way to compute it other than the method you already have.
     
  5. Dec 11, 2009 #4

    Gib Z

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    There is;

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

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

    [tex]= \int \frac{ d(u-\frac{1}{u})}{ (u-\frac{1}{u})^2 +2} du + \int \frac{ d(u+\frac{1}{u})}{ (u+\frac{1}{u})^2 -2} du[/tex]
     
  6. Dec 12, 2009 #5
    I'm sorry, but I don't see exactly how that helps. :confused:
     
  7. Dec 12, 2009 #6

    Gib Z

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    The remaining integrals are of forms which are generally considered standard integrals;

    [tex]\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C[/tex]

    [tex]\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log_e \left( \frac{x-a}{x+a}\right) + C[/tex]
     
  8. Dec 13, 2009 #7
    But how do I deal with the [itex]d\left(u-\frac{1}{u}\right)[/itex] and [itex]d\left(u+\frac{1}{u}\right)[/itex] in the numerators of the integrals?
     
  9. Dec 13, 2009 #8
    Substitution: put z = u-1/u in the first integral and z = u+1/u in the second integral
     
  10. Dec 13, 2009 #9
    murshid_islam - Just to be clear, the trailing [itex]du[/itex] in Gib Z's brillant integrals should be deleted. The form of the integrals are really

    [tex]\int \frac{2u^2}{u^4+1} du = \int \frac{ d(u-\frac{1}{u})}{ (u-\frac{1}{u})^2 +2} + \int \frac{ d(u+\frac{1}{u})}{ (u+\frac{1}{u})^2 -2} [/tex]

    Do you now understand how the substitution gets to the integral forms in his second post?
     
  11. Dec 13, 2009 #10
    Thanks a lot. Now everything is clear to me.


    Oh yes, now I understand.
     
  12. Dec 13, 2009 #11

    Gib Z

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    Argh! All that latex code makes me forget what I'm actually typing and it's just instict to jab a du in there after an integral! Sorry murshid_islam!
     
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