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Math Help - Integration by substitution

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

    I was checking my work against the mathematica integral calculator, and my answer on this problem is totaly different than theirs. Can anyone tell me what I did wrong?

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

    I did

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

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

    \int\frac{xdx}{\sqrt{1-x^{2}}}=\frac{-1}{2}\ln|\sqrt{1-x^{2}}|+c

    mathmatica gives an answer of

    -\sqrt{1-x^{2}}

    I tries using u=\sqrt{1-x^{2}} , but that did not produce an equation I could integrate. Does anyone have any suggestions?
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  2. #2
    Senior Member
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    You don't integrate \frac {1}{\sqrt u} like that, you integrate it normally, so

    \int \frac {1}{\sqrt u}du

    \int u^{-\frac {1}{2}}du

    2u^{\frac {1}{2}}+c
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  3. #3
    Member sinewave85's Avatar
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    Quote Originally Posted by chengbin View Post
    You don't integrate \frac {1}{\sqrt u} like that, you integrate it normally, so

    \int \frac {1}{\sqrt u}du

    \int u^{-\frac {1}{2}}du

    2u^{\frac {1}{2}}+c
    Do'h. Thanks very much!
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