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How to integrate sqrt( cos(x) + cos(x)^3)

  1. Mar 10, 2005 #1
    I need to integrate Sqrt[Cos(x)+(Cos(x))^3]
    Mathematica gives answer with AppelF, I've never heard about it. This excercize was given in my book, so there must be answer in elementary functions. I tryed 1) tg(x/2)=t and 2) Cosx=t , but both didn't help me. Give me please any advice.
    What does AppelF mean?
     
  2. jcsd
  3. Mar 10, 2005 #2

    arildno

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    In all probability, "Appel function".
    You might try to find out more about it on wikipedia.
     
  4. Mar 10, 2005 #3

    Zurtex

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    I would suggest you double check the question, mathematica will just plain not compute that for me.
     
  5. Mar 10, 2005 #4

    dextercioby

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    Are u sure it's
    [tex] \int \sqrt{\cos x +\cos^{3}x} \ dx [/tex]...?

    Daniel.

    EDIT:My Maple is helpless.It reminds me of another integral...
    [tex] \int e^{\sin x} dx [/tex]
     
    Last edited: Mar 10, 2005
  6. Mar 10, 2005 #5

    Zurtex

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    An obvious thought but are you sure it is not:

    [tex]\int \sqrt{\cos (x) - {\cos^3 (x)}} dx[/tex]

    As this then becomes relatively simple.
     
  7. Mar 10, 2005 #6
    Yes, Daniel, You are right. In fact it was definite integral, but I don't remember borders x1 and x2. I think it doesn't matter.
     
  8. Mar 10, 2005 #7

    arildno

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    I think Zurtex has found the correct integrand ..
     
  9. Mar 10, 2005 #8

    dextercioby

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    It doesn't.Wolfram Mathematica's Integrator is helpless,too.

    I'll look it up in G&R,though i doubt it is there.

    Daniel.
     
  10. Mar 10, 2005 #9

    dextercioby

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    U can't apply the FTC,because you can't find the antiderivative.

    Daniel.
     
  11. Mar 10, 2005 #10

    arildno

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    Eeh, assuming Zurtex is right, the anti-derivative [tex]\int\sqrt{\cos{x}-\cos^{3}x}dx=\int|\sin{x}|\sqrt{\cos{x}}dx[/tex] is easily expressed in terms of elementary functions.
     
  12. Mar 10, 2005 #11

    dextercioby

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    Yes,i was OBVIOUSLY referring to the OP's problem,not to an invented one...:wink:

    Daniel.
     
  13. Mar 10, 2005 #12

    arildno

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    Isn't the OP's problem invented??
    Surely, it wasn't discovered..:confused:, :wink:
     
  14. Mar 10, 2005 #13

    dextercioby

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    :wink: For him it was a discovery...:wink:

    He'll remember it as the integral that f***** up all mathematical software...:tongue2:

    Daniel.
     
  15. Mar 10, 2005 #14
    [tex]\sqrt{\cos x} \sqrt{1 + \cos^2x}[/tex]
    How about using t² = cos² and then calculate


    [tex]\int \frac{-t \sqrt{1+t^2}}{\sqrt{1-t^2}} dt[/tex]

    Then integration by parts

    [tex]\frac{-t}{\sqrt{1-t^2}} = d{\sqrt{1-t^2}}[/tex]

    For what it is worth

    marlon
     
    Last edited: Mar 10, 2005
  16. Mar 10, 2005 #15

    dextercioby

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    Your substitution is something like

    [tex]|t|=|\cos x| [/tex] which is very difficult to manipulate.

    Besides,the function under the sq.root can take both positive & negative values.And that's why a substitution must be made with care...

    Daniel.
     
  17. Mar 11, 2005 #16

    saltydog

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    Yegor, can you report the solution technique? What section in the Calculus text did this integral appear? That might give a clue as to how the text wished it to be solved.
     
  18. Mar 11, 2005 #17

    dextercioby

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    Salty,i'm 99% sure that it wasn't that integral he was suppossed to solve...

    Daniel.
     
  19. Mar 11, 2005 #18
    The integration by parts yields :

    [tex] \int \sqrt{1+ t^2} d \sqrt{1-t^2}[/tex]

    [tex]\sqrt{1-t^2} \sqrt{1+t^2} - \int \frac{t \sqrt{1-t^2}}{\sqrt{1+t^2}} = \int \frac{-t \sqrt{1+t^2}}{\sqrt{1-t^2}} dt[/tex]


    Then add up the two integrals...
    [tex] \int \frac{t \sqrt{1-t^2}}{\sqrt{1+t^2}} - \int \frac{t \sqrt{1+t^2}}{\sqrt{1-t^2}}[/tex]

    This sum yields

    [tex]\int \frac{-2t^3 }{\sqrt {1-t^4}}[/tex]

    This is easy to solve

    Indeed keep in mind that with the substitutions you need to be sure the sign is + otherwise they are invalid. Now you can incorporate negative numbers by writing a minus and this works, i checked it out

    marlon
     
    Last edited: Mar 11, 2005
  20. Mar 11, 2005 #19

    dextercioby

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    Marlon,it's chasing your tail...You get nowhere,trust me.

    Daniel.
     
  21. Mar 11, 2005 #20

    dexter don't be so jalous... Prove it...Prove me wrong...

    marlon
     
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