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Math Help - Integral by substitution

  1. #1
    Member sinewave85's Avatar
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    Integration by substitution

    I am stuck!

    \int\frac{dx}{\sqrt{1-2x^{2}}}

    \mbox{Let }u=1-2x^{2}, du=-4xdx

    \int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-1}{4}\int\frac{u^{-1/2}du}{x}

    \int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-1}{4}\int\frac{du}{u^{1/2}(1/2-u/2)^{1/2}}

    \int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-1}{4}\int\frac{du}{(u/2-u^{2}/2)^{1/2}}

    \int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-\sqrt{2}}{4}\int\frac{du}{(u-u^{2})^{1/2}}


    I don't know where to go from there. Any help is greatly appreciated.
    Last edited by sinewave85; August 26th 2009 at 11:53 AM.
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  2. #2
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    Go to this website Wolfram|Alpha
    In the input window, type in this exact expression: integrate 1/sqrt[1-2x^2]dx (you can copy & paste)
    Click the equals bar at the right-hand end of the input window.
    Be sure you click ‘show steps’ to see the solution
    .
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  3. #3
    Member sinewave85's Avatar
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    Quote Originally Posted by Plato View Post
    Go to this website Wolfram|Alpha
    In the input window, type in this exact expression: integrate 1/sqrt[1-2x^2]dx (you can copy & paste)
    Click the equals bar at the right-hand end of the input window.
    Be sure you click ‘show steps’ to see the solution
    .
    Thanks very much for the link. The problem is that it shows the answer, but even the "show steps" window does not really show how to get there. Could you explain it a little bit more?

    I realize that this is a basic integral:

    \int\frac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}(x) + c

    but is that enough to get from this problem to the answer they give without any intermediate steps?

    \int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{sin^{-1}(x\sqrt{2})}{\sqrt{2}}+c
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  4. #4
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    u=\sqrt{2}x
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  5. #5
    Member sinewave85's Avatar
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    Quote Originally Posted by Plato View Post
    u=\sqrt{2}x
    I see (now that you have positioned me directly in front of the solution, that is)! Thanks very much.
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