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Math Help - By making the substitution..

  1. #1
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    By making the substitution..

    By making the substitution u-(sin)^(-1) (x/3), simplify: x^2/√(9-x^2 )
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: By making the substitution..

    Is the substitution:
    u=\sin^{-1}\cdot \frac{x}{3} or?
    Because that would be strange.
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  3. #3
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    Re: By making the substitution..

    \displaystyle u=\sin^{-1} \frac{x}{3}

    \displaystyle \sin u= \frac{x}{3}

    \displaystyle x = 3\sin u

    So the expression becomes

    \displaystyle \frac{x^2}{\sqrt{9-x^2}} = \frac{9\sin^2 u}{\sqrt{9-(3\sin u)^2}} = \dots

    Can you finish it from here?

    Do you need to integrate this function? If so, consider \displaystyle x = 3\sin u \implies dx = 3\cos u~du
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  4. #4
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    Re: By making the substitution..

    yep thats it
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  5. #5
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    Re: By making the substitution..

    Quote Originally Posted by pickslides View Post
    \displaystyle u=\sin^{-1} \frac{x}{3}

    \displaystyle \sin u= \frac{x}{3}

    \displaystyle x = 3\sin u

    So the expression becomes

    \displaystyle \frac{x^2}{\sqrt{9-x^2}} = \frac{9\sin^2 u}{\sqrt{9-(3\sin u)^2}} = \dots

    Can you finish it from here?
    huh im not sure.
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  6. #6
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    Re: By making the substitution..

    Quote Originally Posted by JamesBond16 View Post
    huh im not sure.
    Give it a shot yourself, start by cleaning up the denominator.

    Then take out any common factors inside the radical, then apply the identity \displaystyle \sin^2 x +\cos^2 x = 1 to eliminate the radical.
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  7. #7
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    Re: By making the substitution..

    [3 sin(u)]^2 / sqrt(9 - [3 sin(u)]^2)
    [ 9 sin^2(u) ] / sqrt(9 - 9 sin^2(u))
    [ 9 sin^2(u) ] / sqrt(9{1 - sin^2(u)})
    [ 9 sin^2(u) ] / sqrt(9cos^2(u))
    [ 9 sin^2(u) ] / [3 cos(u)]
    [ 3 sin^2(u) ] / [ cos(u)]
    3 tan(u) sin(u)

    YEa?
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  8. #8
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    Re: By making the substitution..

    Nice work old mate!
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