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Thread: 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?

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

    I did

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

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

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

    mathmatica gives an answer of

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

    I tries using $\displaystyle 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 $\displaystyle \frac {1}{\sqrt u}$ like that, you integrate it normally, so

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

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

    $\displaystyle 2u^{\frac {1}{2}}+c$
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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 $\displaystyle \frac {1}{\sqrt u}$ like that, you integrate it normally, so

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

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

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