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Integral question

  1. Apr 17, 2007 #1
    i added a file with of the integral
    and how i "solved" it

    on the final stage of my integral i gut stuck on a
    complicated fuction

    is there a way to change it so it will look simpler

    plz help
     

    Attached Files:

    • tan4.GIF
      tan4.GIF
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  2. jcsd
  3. Apr 17, 2007 #2

    StatusX

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    Homework Helper

    So you're asking if you can simplify [itex]\sin(2\tan^{-1}(x))[/itex]? Use the double angle formula to get it in terms of [itex]\sin(\tan^{-1}(x))[/itex] and [itex]\cos(\tan^{-1}(x))[/itex]. Then if you put [itex]u=\sin(\tan^{-1}(x))[/itex], you have [itex]\cos(\tan^{-1}(x))=\sqrt{1-u^2}[/itex] and:

    [tex]\frac{u}{\sqrt{1-u^2}}=\frac{\sin(\tan^{-1}(x)}{\cos(\tan^{-1}(x))}=\tan(\tan^{-1}(x))=x,[/tex]

    which you can solve for in terms of u.
     
  4. Apr 18, 2007 #3
    you said u=sin(arctan(x))
    how come i have cos(arctan(x))=(1-u^2)^0.5

    i didnt understand what to do step by step

    ???
     
  5. Apr 18, 2007 #4

    StatusX

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    Because cos^2+sin^2=1.
     
  6. Feb 11, 2009 #5
    but the "u" expression represents x

    i heard of some triangle method of solving stuff like
    [itex]
    \sin(2\tan^{-1}(x))
    [/itex]

    ??
     
  7. Feb 11, 2009 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, u= sin(arctan(x))

    For something simple, like sin(tan-1(x)), you can imagine a triangle with "opposite side" x and "near side" 1 so that "opposite side over near side" = tan(angle)= x and angle= tan-1(x). Then sin(tan-1(x))= sin(angle) which is "near side over hypotenuse". Use the Pythagorean theorem to find the length of the hypotenus: [itex]\sqrt{x^2+ 1}[/itex] and you get sin(tan-1(x))= [itex]x/\sqrt{x^2+ 1}[/itex].

    However, the "2" multiplying tan-1(x) makes that much harder.
     
  8. Feb 11, 2009 #7
    how to build the new integral
    i need to build a du for that which i the derivative of
    u= sin(arctan(x))

    what is the full new integral?
     
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