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Math Help - How to integrate this?

  1. #1
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    How to integrate this?

    Question is in orange, answer is in black.

    I got no idea how they got this answer :\

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

    Just the one question i cant get my head aroung in this exercise step by step method would be appreciated
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  2. #2
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    Let x=\sqrt{\frac{g}{k}}sec{\theta}, \;\ dx=\sqrt{\frac{g}{k}}sec{\theta}tan{\theta}d{\thet  a}

    and it should whittle down nicely.
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  3. #3
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    Quote Originally Posted by Kevlar View Post
    Question is in orange, answer is in black.

    I got no idea how they got this answer :\

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

    Just the one question i cant get my head aroung in this exercise step by step method would be appreciated
    first lets pull some stuff out to get

    -\frac{1}{k} \int \frac{dx}{x^2-\frac{g}{k}}

    For simplicity let \frac{g}{k}=m^2 to get

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

    Using partial fractions we get

    \int \frac{dx}{x^2-m^2}=\int \frac{\frac{1}{2m}}{x-m}dx+\frac{-\frac{1}{2m}}{x+m}dx=

    \frac{1}{2m}\ln|x-m|-\frac{1}{2m}\ln|x+m|=\frac{1}{2m}\ln \left| \frac{x-m}{x+m}\right|=
    -\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)

    So finally we get

    \frac{1}{km}\tanh^{-1}\left( \frac{x}{m}\right)=\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}x \right)
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  4. #4
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    Quote Originally Posted by galactus View Post
    Let x=\sqrt{\frac{g}{k}}sec{\theta}, \;\ dx=\sqrt{\frac{g}{k}}sec{\theta}tan{\theta}d{\thet  a}

    and it should whittle down nicely.
    still got no idea this just completly confused me
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    first lets pull some stuff out to get

    -\frac{1}{k} \int \frac{dx}{x^2-\frac{g}{k}}

    For simplicity let \frac{g}{k}=m^2 to get

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

    Using partial fractions we get

    \int \frac{dx}{x^2-m^2}=\int \frac{\frac{1}{2m}}{x-m}dx+\frac{-\frac{1}{2m}}{x+m}dx=

    \frac{1}{2m}\ln|x-m|-\frac{1}{2m}\ln|x+m|=\frac{1}{2m}\ln \left| \frac{x-m}{x+m}\right|=
    -\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)

    So finally we get

    \frac{1}{km}\tanh^{-1}\left( \frac{x}{m}\right)=\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}x \right)
    How do you get from the ln to inverse tanh?
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  6. #6
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    Quote Originally Posted by Kevlar View Post
    still got no idea this just completly confused me
    That's just trig substitution. EmptySet has a nicer solution.

    Note the identity tanh^{-1}(x)=\frac{1}{2}ln(\frac{1+x}{1-x})

    But, note similarity with \int\frac{1}{a^{2}+u^{2}}du=\frac{1}{a}tan^{-1}(\frac{u}{a})+C
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