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Math Help - Trig substitution

  1. #1
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    Trig substitution

    (sqrt(9-x^2))/(x^2)

    Evaluate the intergral.... please
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  2. #2
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    Quote Originally Posted by tiga killa View Post
    (sqrt(9-x^2))/(x^2)

    Evaluate the intergral.... please
    Let x=3\sin(\theta) \implies dx=3\cos(\theta) d\theta

    \int \frac{\sqrt{9-9\sin^2(\theta)}(3\cos(\theta)}{9\sin^2(x)}d\theta  =\int\frac{\cos^2(\theta)}{\sin^2(\theta)}d\theta

    \int \frac{1-\sin^2(\theta)}{\sin^2(\theta)}d\theta=\int \csc^2(\theta)d\theta-\int d\theta=-\cot(\theta)-\theta

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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by tiga killa View Post
    (sqrt(9-x^2))/(x^2)

    Evaluate the intergral.... please
    Let x=3\sin u. Thus dx=3\cos u\,du

    Now the integral is:
    \int\frac{\sqrt{9-9\sin^2u}}{9\sin^2u}\cdot3\cos u\,du = \int\frac{\cos^2 u}{\sin^2 u}\,du = \int\frac{1-\sin^2u}{\sin^2u}\,du = \int\csc^2u\,du - \int\,du = -\cot u -u

    Sub back in to get -\cot\left(\sin^{-1}\left(\frac{x}{3}\right)\right)-\sin^{-1}\left(\frac{x}{3}\right) = \boxed{-\frac{\sqrt{9-x^2}}{x}-\sin^{-1}\left(\frac{x}{3}\right)}

    EDIT: Beat to it by less than a minute!
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