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Thread: stuck with sin^4(x)

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    52

    stuck with sin^4(x)

    Here is what I am trying to do:

    $\displaystyle

    \int{\sin^4{x}} = \int{\sin^3{x}\sin{x}}
    $
    now use product rule
    $\displaystyle
    \int{udv} = uv - \int{vdu}

    $

    notably

    $\displaystyle

    dv = \sin^3{x}
    v = \frac{\cos^3{x}}{3} - \cos{x}
    $

    I derived this earlier and just copied the result. Now am I supposed to do :

    $\displaystyle
    \int{vdu}
    $

    Isn't that just

    $\displaystyle
    (\frac{\cos^3{x}}{3} - \cos{x})\cos{x}
    $

    I feel like I am going backwards...
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  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
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    North Texas
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    $\displaystyle \sin^4{x} = $

    $\displaystyle (\sin^2{x})^2 = $

    $\displaystyle \left[\frac{1 - \cos(2x)}{2}\right]^2 =$

    $\displaystyle \frac{1}{4}\left[1 - 2\cos(2x) + \cos^2(2x)\right] =$

    $\displaystyle \frac{1}{4}\left[1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}\right] =
    $

    $\displaystyle \frac{3}{8} - \frac{\cos(2x)}{2} + \frac{\cos(4x)}{8}$

    now integrate
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