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anonymous

  • 5 years ago

integral dx over square root of quantity of 2x-x^2

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  1. anonymous
    • 5 years ago
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    Focus on the quantity within the square root You can rewrite 2x -x^2 as 1- (x^2 -2x +1) this gives you \[\int\limits_{}^{}dx / (\sqrt{1-(x-1)^{2}}\] now put x-1=sin t dx=cost dt the above expression converts to costdt/[cost] =\[\int\limits_{}^{}\]dt =t =\[\sin^{-1} (x-1)\]

  2. anonymous
    • 5 years ago
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    you are my hero!!!

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