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Math Help - Dose this integration seem right

  1. #1
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    Dose this integration seem right

    The question h(u)=sin^2(8u)
    I used the double angle formulas and the answer I got is
    1/32sin(16u)+1/2u+c
    Is this right
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  2. #2
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    Re: Dose this integration seem right

    Hello, duggielanger!

    \int \sin^2(8u)\,du

    I used the double angle formulas and got: . \tfrac{1}{32}\sin(16u)+\tfrac{1}{2}u+C
    Is this right? . no

    If you differentiate your answer, you do not get \sin^2(8u)


    \int\sin^2(8y)\,du \;=\;\tfrac{1}{2}\int\big[1 - \cos(16u)\big]\,du

    . . . . . . . . . . =\;\tfrac{1}{2}\left[u - \tfrac{1}{16}\sin(16u)\right] + C

    . . . . . . . . . . =\;\tfrac{1}{2}u - \tfrac{1}{32}\sin(16u) + C
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  3. #3
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    Re: Dose this integration seem right

    Ok i can't see where I've gone wrong in my workings out so I will post them and could you have a look please (can't find a theta sign so i'll type
    it)
    sin^2(8u)
    where
    sin^2=1-cos^2theta here theta=8u
    so
    sin^2(8u)=1-cos^2(8u)
    sin^2(8u)=1-cos^2(16u)
    sin^2(8u)=1/2(cos(16u)+1)
    sin^2(8u)=(1/2cos(16u)+1/2)=1/32sin(16u)-1/2u+c

    Thank you
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