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Indefinite Integration Problem

  1. Jul 23, 2005 #1


    How is this problem integrated?

    [tex]\int \sqrt{ \sin x} \; dx[/tex]

     
  2. jcsd
  3. Jul 23, 2005 #2

    Zurtex

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    The solution can not be given in terms of elementary functions.
     
  4. Jul 23, 2005 #3

    lurflurf

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    No elementary functions have [tex]\sqrt{\sin(x)}[/tex] as their derivative. It looks like it an antiderivative could be expressed using elliptic integral of the second kind.
    http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html
     
    Last edited: Jul 23, 2005
  5. Jul 23, 2005 #4

    Zurtex

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    Indeed, according to mathematica where:

    [tex]\text{EllipticE} (z,m) = \int_0^z \sqrt{1 - m \sin^2 t} \, dt[/tex]

    Then:

    [tex]\int \sqrt{ \sin x } \, dx = -2 \text{EllipticE} \left( \frac{1}{4} ( \pi - 2x ), 2 \right)[/tex]
     
  6. Jul 23, 2005 #5

    saltydog

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    Alright, I had problems with it:

    Show:

    [tex]\int \sqrt{ \sin x } \, dx = -2 \text{EllipticE} \left( \frac{1}{4} ( \pi - 2x ), 2 \right)[/tex]

    So differentiating the RHS:

    [tex]\frac{d}{dx}\left(-2\int_0^{\frac{\pi-2x}{4}} \sqrt{1-2Sin^2(t)}dt\right)=\sqrt{1-2Sin^2(\frac{\pi}{4}-\frac{x}{2})}[/tex]

    Well anyway, I'll save the rest for others to go through if they need practice like me to show that this is equal to:

    [tex]\sqrt{\sin x}[/tex]
     
    Last edited: Jul 23, 2005
  7. Jul 23, 2005 #6

    Zurtex

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    :biggrin: I had the same the same odd feeling about it when I looked at it and enjoyed doing my old A-Level work on proving trigonometric identities.
     
  8. Jul 24, 2005 #7
    Integral Nexus...



    Integrating with EllipticE formula:
    [tex]\int \sqrt{\cos x} \; dx = 2 \int_0^{\frac{x}{2}} \sqrt{1 - 2 \sin^2 (t)} \; dt[/tex]

    RHS Differentiation:
    [tex]\frac{d}{dx} \left( 2 \int_0^{\frac{x}{2}} \sqrt{1 - 2 \sin^2 (t)} \; dt \right) = \sqrt{1 - 2 \sin^2 \left( \frac{x}{2} \right)}[/tex]

    Trigonometric Identity:
    [tex]1 - 2 \sin^2 \left( \frac{x}{2} \right) = \cos x[/tex]

    Any Calculus I students interested in integrating this formula?
    [tex]\int \sqrt{ \tan x} \; dx[/tex]

     
  9. Jul 24, 2005 #8

    Zurtex

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    Haha, that's really nasty, if anyone wants a go at this then I'll just help out by saying that you don't need to put it of the form of an EllipticE formula
     
  10. Jul 24, 2005 #9

    lurflurf

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    It is not that bad. It should probably be on a list of good calculus final questions along with.
    [tex]\frac{d}{dx}x^x[/tex]
    and
    [tex]\int x^3e^{-2x}\sin(x)dx[/tex]
     
  11. Jul 24, 2005 #10

    Zurtex

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    I got asked that question on my university interview, one of the few people who did it without any help :approve:
     
  12. Jul 24, 2005 #11

    TD

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    The following substitution
    [tex]\tan x = y^2 \Leftrightarrow x = \arctan \left( {y^2 } \right) \Leftrightarrow dx = \frac{{2y}}{{y^4 + 1}}dy[/tex]
    gives: [tex]\int {\frac{{2y^2 }}{{y^4 + 1}}dy}[/tex]

    That should be doable :smile:
     
  13. Jul 24, 2005 #12

    lurflurf

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    The only potential problems is that if one "refuses" to use complex numbers, effecting the integration requires some rather unmotivated ad hoc manipulations. Probably some rather clever trig identitiy manipulation would get the job done as well.
     
    Last edited: Jul 24, 2005
  14. Jul 26, 2005 #13
    Texas Nexus...



    Have any Calculus I students attempted to compute this formula on a >TI-89 calculator?

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

    How long did the computation take? stopwatch?

    Now compute this formula:
    [tex]\int {\frac{{2y^2 }}{{y^4 + 1}}dy}[/tex]

    How long did the computation take?

    Have any Calculus I students compared the tangent formula solution generated from a >TI-89 as compared to the Mathematica solution?

    Reference:
    http://integrals.wolfram.com/

     
    Last edited: Jul 26, 2005
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