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Math Help - integration no.4

  1. #1
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    integration no.4

    revise: posting a wrong question



    Last edited by afeasfaerw23231233; October 27th 2007 at 08:35 AM.
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  2. #2
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    Just use trig sub.

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

    Let x=sin{\theta}, \;\ dx=cos{\theta}d{\theta}

    Make the subs and it simplifies down to:

    \int{cot^{2}{\theta}}d{\theta}


    \int{cot^{n}u}du=\frac{-1}{n-1}cot^{n-1}u-\int{cot^{n-2}u}du

    After you integrate cot, then make the sub {\theta}=sin^{-1}(x) to get it back in terms of x.

    You're finished.
    Last edited by galactus; October 27th 2007 at 04:53 AM.
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  3. #3
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    simple answer: coz you did it wrong.
    split it up first: √[1/x^4 + 1/x] dx
    then do the substitution, u = 1/x, dx = -1/u du

    -∫√[u^4 + u] * 1/u du

    factor out u from inside the square root and cancel one from denominator, you get
    -∫√[u-1]/u du
    you can use a trig substitution to solve this.
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  4. #4
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    revise: posting a wrong question
    Attached Thumbnails Attached Thumbnails integration no.4-revise.gif  
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  5. #5
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    Krizalid's Avatar
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    Does this make sense?
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