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Math Help - indefinite integrals

  1. #1
    Junior Member
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    indefinite integrals

    Hi Everyone,

    Just to make sure I did it right, everytime I'm ok a differente rule arise

    Here is my Question and answer

    Find the indefinite integrals of the following functions.




    (i)h(u) = cos2(3u)

    ∫ cos2(3u) du

    ∫ 1/3 sin2 (3u) = 1/3 sin2 + 3u + c



    Thanks again for helping me
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by valerie-petit View Post
    Hi Everyone,

    Just to make sure I did it right, everytime I'm ok a differente rule arise

    Here is my Question and answer

    Find the indefinite integrals of the following functions.




    (i)h(u) = cos2(3u)

    ∫ cos2(3u) du

    ∫ 1/3 sin2 (3u) = 1/3 sin2 + 3u + c



    Thanks again for helping me
    Sigh...

    I assume this is

    \int\cos^2(3u)du

    and not

    \cos(3(2u))du

    FYI holding the alt key, while holding it press 0178 and the realease it, it will create ²...0179 gives ³.

    But anyways

    \cos^2(3u)=\frac{1+\cos(6u)}{2}

    So we have

    \frac{1}{2}\int\bigg[1+\cos(6u\bigg]du=\frac{1}{2}\bigg[u+\frac{1}{6}\sin(6u)\bigg]+C
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  3. #3
    Junior Member
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    Thanks again Mathstud28 for your help, you right it is cos²(3u)du

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