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Math Help - Easy integration problem.

  1. #1
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    Easy integration problem.



    Thanks.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by usagi_killer View Post


    Thanks.

    let x=3\sin(t) \implies dx=3\cos(t)dt

    \int \sqrt{9-x^2}dx=\int \sqrt{9-9\sin^2(t)}(3\cos(t))dt=

    9\int \cos^2(t)dt =\frac{9}{2}\int 1+\cos(2t)dt=\frac{9}{2}t+\frac{9}{4}\sin(2t)+c=

    \frac{9}{2}t+\frac{9}{2}\sin(t)\cos(t)+C

    Now we need to solve for t and cos(t)

    x=3\sin(t) \iff \frac{x}{3}=\sin(t) \iff \frac{x^2}{9}=\sin^2(t) \iff \frac{x^2}{9}=1-\cos^2(t)

     \iff \cos^2(t)=1-\frac{x^2}{9} \iff \cos(t)=\frac{\sqrt{9-x^2}}{3}

    x=3\sin(t) \iff \sin^{-1}\left( \frac{x}{3}\right)=t

    \frac{9}{2}\sin^{-1}\left( \frac{x}{3}\right)+\frac{9}{2}\frac{x\sqrt{9-x^2}}{9}+C

    \frac{9}{2}\sin^{-1}\left( \frac{x}{3}\right)+\frac{x\sqrt{9-x^2}}{2}+C
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  3. #3
    Senior Member Twig's Avatar
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    Let  \frac{x}{3} = sin(t)

     \int \sqrt{9-x^{2}} dx = \int 3 \sqrt{1-\frac{x^{2}}{9}} = 3 \int \sqrt{1-sin^{2}(t)} \cdot 3 cos(t) dt \Rightarrow

     9 \int cos^{2}(t) dt

    This can be re-written using the double angle formula  cos(2t)=2cos^{2}(t)-1

     \frac{9}{2} \int (1+cos(2t)) dt = \frac{9}{2}(t+\frac{1}{2}sin(2t)) + C


    I gotten beaten to it here, the above solution is better! I just realized I didnt solve for x in the end.. =)
    Last edited by Twig; June 13th 2009 at 08:28 AM. Reason: See above...
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by usagi_killer View Post


    Thanks.
    let 3sinu =x ...... (3cosu) du = dx

    from assumptions 3sinu=x we can find cosu

    sin^2u + cos^2u =1 but sinu =\frac{x}{3}

    \frac{x^2}{9} + cos^2u = 1

    cosu = \frac{\sqrt{9-x^2}}{3}

    \int \sqrt{9-x^2} dx

    \int {\color{red}\sqrt{9-x^2}} \left(\frac{3}{{\color{red}3}}\right) (3cosu) du

    the red terms is cosu

    \int (3cos^2u) du you can solve it right .
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