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Math Help - stuck with sin^4(x)

  1. #1
    Junior Member
    Joined
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    stuck with sin^4(x)

    Here is what I am trying to do:

    <br /> <br />
\int{\sin^4{x}} = \int{\sin^3{x}\sin{x}}<br />
    now use product rule
    <br />
\int{udv} = uv - \int{vdu}<br /> <br />

    notably

    <br /> <br />
dv = \sin^3{x}<br />
v = \frac{\cos^3{x}}{3} - \cos{x}<br />

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

    <br />
\int{vdu}<br />

    Isn't that just

    <br />
(\frac{\cos^3{x}}{3} - \cos{x})\cos{x}<br />

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

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

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

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

    \frac{1}{4}\left[1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}\right] =<br />

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

    now integrate
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