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Math Help - Integral cos^2(mx)

  1. #1
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    Integral cos^2(mx)

    I'd love to see various ways to solve this integral, please.

    \int_0^{2\pi} \cos^2(mx)\rm{d}x \qquad m\in \mathbb N

    I have thought about using that

    \cos^2(x) = \frac{1+\cos(2x)}{2}

    as I know that m\in \mathbb N, but I'm not sure if it is OK. However, I hope someone feels like showing me various ways of doing it.

    In advance, thank you for your time and help.
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  2. #2
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    u = mx \ \Rightarrow \ du = m dx

    So: \int_{0}^{2\pi} \cos^2 (mx) \ dx = \frac{1}{m} \int_{u(0)}^{u(2\pi)} \cos^2 u \ du

    Now use the identity: \cos^2 x = \frac{1 + \cos 2x}{2}
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  3. #3
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    Quote Originally Posted by MatteNoob View Post
    I'd love to see various ways to solve this integral, please.

    \int_0^{2\pi} \cos^2(mx)\rm{d}x \qquad m\in \mathbb N

    I have thought about using that

    \cos^2(x) = \frac{1+\cos(2x)}{2}

    as I know that m\in \mathbb N, but I'm not sure if it is OK. However, I hope someone feels like showing me various ways of doing it.

    In advance, thank you for your time and help.
    \int_0^{2\pi} \cos^2(mx)\rm{d}x \qquad m\in \mathbb N

    =\int_0^{2\pi} \frac{1+\cos(2mx)}{2}~dx

    =\frac{1}{2}\int_0^{2\pi} \left[ 1+\cos(2mx)\right]~dx
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  4. #4
    Super Member Showcase_22's Avatar
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    I have this way:



    I'd like to point out that I don't usually use such a long and windy method, it's just that you wanted various ways of solving it.

    I just noticed there's a typo on my image. For the second substitution it should have a -x/2m. See my last post.
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  5. #5
    Super Member Showcase_22's Avatar
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    This is another long and windy method:



    Just use the cos^2(x)+sin^2(x)=1 and plug in some numbers to finish it off.
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  6. #6
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    (sorry about 3 posts in a row, it's just that I can't get the image thing to pop up when I use "edit").

    This is less of an integrating method and more of just rewriting the question:



    One of the above methods would need to be used to solve it though.
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