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Math Help - Integral of t \sqrt (2t - t^2)

  1. #1
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    Integral of t \sqrt (2t - t^2)

    \int {t}{\sqrt{2t-t^2}}dt
    this integral has a long answer.... i need advice on what to do first... then when im stuck on sumthing ill ask for another help here...
    thanks
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  2. #2
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    One recourse may be to let t=u+1, \;\ t-1=u, \;\ du=dt

    This gives:

    \int\sqrt{1-u^{2}}(u+1)du

    Now, perhaps try a trig sub.

    Just a quick thought from glancing at it.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    \int {t}{\sqrt{2t-t^2}}dt
    this integral has a long answer.... i need advice on what to do first... then when im stuck on sumthing ill ask for another help here...
    thanks
    Complete the square in the radical:
    \int {t}{\sqrt{2t - t^2}}dt

    = \int t\sqrt{1 + (-1 + 2t - t^2)}dt

    = \int t\sqrt{1 - (1 - 2t + t^2)}dt

    = \int t\sqrt{1 - (t - 1)^2}dt

    Now let x = t - 1 \implies dx = dt

    = \int (x + 1)\sqrt{1 - x^2}dx

    = \int x\sqrt{1 - x^2}dx + \int \sqrt{1 - x^2} dx

    Try these.

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    Complete the square in the radical:
    \int {t}{\sqrt{2t - t^2}}dt

    = \int t\sqrt{1 + (-1 + 2t - t^2)}dt

    = \int t\sqrt{1 - (1 - 2t + t^2)}dt

    = \int t\sqrt{1 - (t - 1)^2}dt
    Then
    Attached Thumbnails Attached Thumbnails Integral of t \sqrt (2t - t^2)-aug1.gif  
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