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Math Help - simple Integral problem

  1. #1
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    simple Integral problem

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

    I know this is related to Inverse Tringemetric functions, so the Integral should somehow end {sin^{-1}}x
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  2. #2
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    try x = 3 \sin u
    Bobak
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  3. #3
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    Sorry, are you meaning using x = 3 \sin u for u substitution?
    \int \frac{{\cos u}}{\sqrt{9-3 \sin u}}
    I can't see how that works :s
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  4. #4
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    Quote Originally Posted by hlolli View Post
    Sorry, are you meaning using x = 3 \sin u for u substitution?
    \int \frac{{\cos u}}{\sqrt{9-3 \sin u}}
    I can't see how that works :s
    you made a little error. you didn't square x, also  dx = 3 \cos u du

    Bobak
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  5. #5
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    \int \frac{3{\cos u}du}{\sqrt{9-9 \sin^{2}u}}

    \int \frac{1}{\sqrt{9-9 \sin^{2}u}}{3{\cos u}du}

    But please stay with me, this is in a chapter of Tringometric functions

    and
    \frac{d}{dx} (\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}

    This should somehow relate, now I'm stuck with 9 that I want to become 1
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  6. #6
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    9-9 \sin^{2}u = 9(1-\sin^{2}u ) = ....

    Bobak
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  7. #7
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    does sin^2 x + cos^2 x = 1 ring any bells?
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  8. #8
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    It's solved Thanks!
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