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Funny integral

  1. Oct 2, 2005 #1
    [SOLVED] funny integral

    hi everyone

    its funny but all maths-software fail solving this "simple" integral

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

    do you know another funny integrals?
     
  2. jcsd
  3. Oct 2, 2005 #2
    Mathematica gives an answer:

    [tex]\frac{-2 \tan ^{-1}\left(1-\sqrt{2} \tan ^{\frac{1}{2}}(x)\right)+2 \tan ^{-1}\left(\sqrt{2} \tan ^{\frac{1}{2}}(x)+1\right)+\log
    \left(-\tan (x)+\sqrt{2} \tan ^{\frac{1}{2}}(x)-1\right)-\log \left(\tan (x)+\sqrt{2} \tan ^{\frac{1}{2}}(x)+1\right)}{2
    \sqrt{2}}[/tex]
     
  4. Oct 2, 2005 #3

    lurflurf

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    [tex]\int \sqrt{\tan(x)}dx=\int\frac{\sec^2(x)}{1+\tan^2(x)}\sqrt{\tan(x)}dx=\int\frac{2\tan(x)}{1+\tan^2(x)}d\sqrt{\tan(x)}[/tex]
    so it hinges on the always fun
    [tex]\int\frac{x^2}{1+x^4}dx[/tex]
     
  5. Oct 2, 2005 #4
    What is about this integral .. ?

    [tex] \int_0^{\frac \pi 2} \ln ( \sin x ) . \ dx [/tex]

    Can the mathematical softwares, such as Maple amd Mathematica give you
    the answer :[tex] - \frac \pi 2 \ln 2 [/tex] ?
     
    Last edited: Oct 2, 2005
  6. Oct 2, 2005 #5
    Change ln to its integral form (so you get a double integral) and use change of variables on that form.
     
  7. Oct 2, 2005 #6
    Unfortunately, I edited my previous replay after you replied .. !

    I wanted to say that the answer of this integral can't be obtained by Maple or Mathematica ..

    Also , I solved the integral with a method different from your method .
     
    Last edited: Oct 2, 2005
  8. Oct 11, 2005 #7
    Wow, I was looking up ways to figure out how to use the LaTeX graphics so that I type [tex]\int \sqrt{\tan(x)}dx[/tex] and ask for help solving that. It's a funny coincidence that I stumbled into this thread. I'm a very lucky person.

    Anyway... I can transform the integral into [tex]2\int\frac{u^2}{1+u^4}du[/tex]. I know what the antiderivative of that is is (I found it in a book of mathematical tables), but I don't know how to prove it. Do you happen to know how to find the antiderivative of [tex]\frac{u^2}{1+u^4}[/tex]?
     
    Last edited: Oct 11, 2005
  9. Oct 11, 2005 #8
    There was a long thread on this a while back. Can't find it though. Another strange one I've found (similar to one posted above) is:

    [tex]\int\frac{1}{1+x^4}\,dx[/tex]

    You have to use that fact that x4+1=(x2+1)2-2x2. And then use the difference of two squares. It gets messy.

    Alex
     
  10. Oct 12, 2005 #9

    Tide

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    You didn't try Derive 6! :)
     
  11. Oct 12, 2005 #10

    dextercioby

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    Not all, my ancient version of Maple (5.3 i guess) gives

    [tex] \int \sqrt{\tan x}dx=\frac{1}{2}\sqrt{2}\arctan \sqrt{2}\frac{\tan ^{\frac{1}{2}}x}{1-\tan x}-\frac{1}{2}\sqrt{2}\ln \frac{\tan x+\sqrt{2}\tan ^{\frac{1}{2}}x+1}{\sqrt{\left( 1+\tan ^{2}x\right) }} + C[/tex].

    Daniel.
     
  12. Oct 12, 2005 #11

    Tide

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    Gellman,

    Incidentally, MuPAD also gives the correct result. What package were you using?
     
  13. Oct 18, 2005 #12
    This a method to solve the integral ..
    [tex] \int \frac { u^2 } { u^4 +1 } du =\frac 12 \int \frac { 2u^2 } { u^4 +1 } du = \frac 12 \int \frac { u^2 -1 } { u^4 +1 } du + \frac { u^2 +1 } { u^4 +1 } du [/tex]
    [tex] = \frac 12 \int \frac { 1 - \frac 1 {u^2} }{ u^2 + \frac 1{u^2} } du
    +\frac 12 \int \frac { 1 + \frac 1 {u^2} }{ u^2 + \frac 1{u^2} } du
    [/tex]
    [tex]= \frac 12 \int \frac { 1 - \frac 1 {u^2} }{ \left (u + \frac 1u \right ) ^2 -1 } du
    +\frac 12 \int \frac { 1 + \frac 1 {u^2} }{ \left (u - \frac 1u \right)^2 +1 } du [/tex]
    [tex] =\frac 12 \int \frac { d \left (u + \frac 1u \right ) }{ \left (u + \frac 1u \right ) ^2 -1 }
    +\frac 12 \int \frac { d \left (u - \frac 1u \right ) }{ \left (u - \frac 1u \right)^2 +1 } [/tex]

    [tex] \mbox { In the first integral , make the subsitution :} v = u + \frac 1u [/tex]
    [tex] \mbox { and in the second integral , make the subsitution : } v = u - \frac 1u [/tex]

    The integrals become now simple , you can integrate them easily
     
    Last edited: Oct 18, 2005
  14. Oct 20, 2005 #13

    benorin

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    Yes, that fact indeed. Read attached gif, see what is meant.
     

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