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Math Help - Definite integral over a period

  1. #1
    Member Mollier's Avatar
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    Definite integral over a period

    Hi,

    how do I integrate,

    \int^{\pi}_0 (8-8cos(2t))^{1/2}dt ?

    I've tried by defining

    u=8-8cos(2t) and then

    \frac{du}{dt}=16sin(2t) \Rightarrow dt=\frac{1}{16sin(2t)}du,

    but that doesn't get me very far.

    I think I remember that there are some trick when dealing with definite integrals over periods and when the function is trigonometric. I don't remember how it works though..
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  2. #2
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    8-8\cos{2t} = 8(1-\cos{2t}) = 16\sin^2{t}.
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  3. #3
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    Remember that \cos{2t} = 1 - 2\sin^2{t}, so

    \int{\sqrt{8 - 8\cos{2t}}\,dt} = \int{\sqrt{8 - 8(1 - 2\sin^2{t})}\,dt}

     = \int{\sqrt{8 - 8 + 16\sin^2{t}}\,dt}

     = \int{\sqrt{16\sin^2{t}}\,dt}

     = \int{4\sin{t}\,dt}

     = -4\cos{t} + C.


    Therefore

    \int_0^\pi{\sqrt{8 - 8\cos{2t}}\,dt} = \left[-4\cos{t}\right]_0^\pi

     = \left[-4\cos{\pi}\right] - \left[-4\cos{0}\right]

     = -4(-1) + 4(1)

     = 4 + 4

     = 8.
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  4. #4
    Member Mollier's Avatar
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    Ah, all those trig identities!!

    Thank you both very much.
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