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Math Help - Standard Integration

  1. #1
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    Standard Integration

    Q: \int \frac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x

    Is there something really simple that I am not seeing? How would this be done? Thanks in advance.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Air View Post
    Q: \int \frac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x

    Is there something really simple that I am not seeing? How would this be done? Thanks in advance.
    Apply a U-Substitution. u=1-x^2. You can take it from there.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Apply a U-Substitution. u=1-x^2. You can take it from there.
    Hehe...I overlooked it too. You can't use u-substitution...give me a minute to think it over...
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Air View Post
    Q: \int \frac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x

    Is there something really simple that I am not seeing? How would this be done? Thanks in advance.
    Apply a trig substitution x=\sin\theta.

    Also, recall that \int \csc\theta \,d\theta= -\ln|\csc\theta+\cot\theta|+C
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  5. #5
    Moo
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    Hello,

    You can make a substitution t=\sqrt{1-x^2}

    \implies dt=\frac{-x}{t} \ dx \implies dx=-\frac tx \ dt


    \int \frac{1}{x \sqrt{1-x^2}} dx=\int \frac{-1}{x^2} dt=- \int \frac{1}{1-t^2} dt=-\frac 12 \int \frac{1}{1-t}+\frac{1}{1+t} dt




    (modulo my mistakes )
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  6. #6
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    Make the substitution z^2=1-x^2.

    (Moo beat me for just one minute.)
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