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Integrating 1/[1+sq.root(tanx)]

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    How to integrate ∫1/(1+√tanx) dx?

    2. The attempt at a solution

    Tried to substitute √tanx with z, but ultimately getting messed up with large expression
     
  2. jcsd
  3. Nov 7, 2012 #2

    tiny-tim

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    hi justwild! :smile:
    if it's the correct large expression, what's the difficulty? :wink:

    show us what you got :smile:
     
  4. Nov 8, 2012 #3
    we have,
    ∫1/[1+√tanx] dx ----------1

    let tanx=z[itex]^{2}[/itex]

    hence by applying derivative on both sides,

    sec[itex]^{2}[/itex]x dx = 2zdz
    [itex]\Rightarrow[/itex] 1+tan[itex]^{2}[/itex]x dx=2zdz
    [itex]\Rightarrow[/itex] 1+z[itex]^{4}[/itex] dx=2zdz
    therefore, dx = [2z/(1+z[itex]^{4}[/itex])]dz

    Substituting the above value in expression 1 we have,
    ∫2z/[{1+z[itex]^{4}[/itex]}{1+z}] dz
    This is what I got and I could not go on from here
     
  5. Nov 8, 2012 #4

    tiny-tim

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    hi justwild! :wink:

    yes, that looks right :smile:

    now use partial fractions

    (factoring (1 + z4) = (1 + iz2)(1 - iz2) = (1 + eiπ/4)timesetc :wink:)
     
  6. Nov 8, 2012 #5
    That is OK but I can't use complex numbers here...
     
  7. Nov 8, 2012 #6

    SammyS

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    z4 + 1 can be factored into two quadratics.

    [itex]z^4+1=(z^2+\sqrt{2}\,x+1)(z^2-\sqrt{2}\,x+1)[/itex]

    Then try partial fractions.

    The overall result of the integration promises to be pretty complicated.
     
  8. Nov 8, 2012 #7

    Ray Vickson

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    The answer will not be complex; the complex numbers come in conjugate pairs, and when the final answer is obtained the imaginary parts go away. This often happens: we end up with some real solution by going out into the complex plane, doing some manipulations there, and finally arriving back on the real line at the end.

    RGV
     
  9. Nov 9, 2012 #8

    tiny-tim

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    a good hint is to complete the square

    z4 + 1 = (z4 + 2z2 + 1) - 2z2 :wink:

    works for any symmetric quartic:

    z4 + 2az3 + bz2 + 2az + 1

    = (z2 + az + 1)2 - (a2 + 2 - b)z2

    = (z2 + (a + √(a2 + 2 - b)z + 1)(z2 + (a - √(a2 + 2 - b)z + 1)​
     
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