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Need help on integration question i found on net

  1. Sep 26, 2009 #1
    Question is in orange, answer is in black.
    2hhen4i.png
    I got no idea how they got this answer :\

    The way im trying is using a substition of [itex]kx^2 = gsin\theta[/itex]

    Just the one question i cant get my head aroung in this exercise step by step method would be appreciated
     
  2. jcsd
  3. Sep 26, 2009 #2

    tiny-tim

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    Hi Keval! :smile:

    Not sin, but tanh … try substituting x√(k/g) = tanhu :wink:
     
  4. Sep 26, 2009 #3
    With denominator instead k x^2 + g you get an arctan answer, right? Use the same steps on this one and get an atanh answer.
     
  5. Sep 26, 2009 #4
    someone check this out for me ??

    [itex]-\frac{1}{k} \int \frac{dx}{x^2-\frac{g}{k}}[/itex]


    to simplify i let[itex]\frac{g}{k}=m^2[/itex] to get


    [itex]-\frac{1}{k}\int \frac{dx}{x^2-m^2}=-\frac{1}{k}\int \frac{dx}{(x-m)(x+m)}[/itex]


    Using partial fractions i got
    [itex]\int \frac{dx}{x^2-m^2}=\int \frac{\frac{1}{2m}}{x-m}dx+\frac{-\frac{1}{2m}}{x+m}dx=[/itex]



    [itex]\frac{1}{2m}\ln|x-m|-\frac{1}{2m}\ln|x+m|=\frac{1}{2m}\ln \left| \frac{x-m}{x+m}\right|=[/itex]

    [itex]-\frac{1}{m}\cdot \frac{1}{2}\ln \left|\frac{x+m}{x-m} \right|=-\frac{1}{m}\tanh^{-1}\left( \frac{x}{m}\right)[/itex]

    [itex]\frac{1}{km}\tanh^{-1}\left( \frac{x}{m}\right)=\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}x \right)[/itex]
     
  6. Sep 27, 2009 #5

    tiny-tim

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    Hi Keval! :wink:

    Yes, your partial fractions method is fine. :smile:

    But it would be far quicker to start "let x√(k/g) = tanhu, dx√(k/g) = sech2u du" …

    try it! :smile:
     
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