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Math Help - Finding integrals

  1. #1
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    Finding integrals

    find the indefinite integral of (x^2)/[(9-x^2)^(1/2)]
    Im think im supposed to try to set x=3sin(u) thus dx=3cos(u) du i think im on the right track but i just keep getting the integral of 9sin^2(u) du and i get all messed up
    i hope that makes sense. i dont know how to type all crazy math like
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  2. #2
    o_O
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    That's very good. Now use the identity: \sin^2 x = \frac{1 - \cos 2x}{2}

    This is much easier to integrate.
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  3. #3
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    okay then i split it up as 2 integrals like so 9{intieral(1/2)-integral[cos(2u)/2]}
    then the the antiderivative of those come out to 9[(u/2)-(sin(2u)/4)]
    then when i plug arcsin(x/3) into u it gets all messed up and i have to take the sin of the arcsin?
    i dont know if im doing something wrong
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  4. #4
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    Use the double angle identity for Sine:

    \sin{2x} = 2\sin{x}\cos{x}

    Then I'm sure you know how to figure out \sin{x} and \cos{x}
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  5. #5
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    umm i understand that the sin(u) would then equal x/3 but what happens to the cos(u)? i have cos(u)=dx/3?
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  6. #6
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    tan\theta = \frac{\text{opposite}}{\text{adjacent}}



    sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}



    cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}
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  7. #7
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    Where did we get a triangle from?
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  8. #8
    Super Member 11rdc11's Avatar
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    sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}~=~\frac{  x}{3}

    Now using this information find out what cos\theta is.

    Hint use pythargeon theorem to find the adjacent side
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  9. #9
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    finally i see it thank you!!!
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