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Math Help - definite integral!

  1. #1
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    definite integral!

     \int_{-\pi}^{\pi} \cos^{2}ny dy

     \left[\frac{n\cos^{3}ny}{3n}\sin{ny}\right]_{-\pi}^{\pi}

     \left[\frac{n\cos^{3}n\pi}{3n}\sin{n\pi}+\frac{n\cos^{3}  n\pi}{3n}\sin{n\pi}\right]

     2\left[\frac{n\cos^{3}n\pi}{3n}\sin{n\pi}\right]

     2\left[0\right]

    pls is 0 equivalent to  2{\pi} or  \pi

    thanks!
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  2. #2
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    Re: definite integral!

    Quote Originally Posted by lawochekel View Post
     \int_{-\pi}^{\pi} \cos^{2}ny dy

     \left[\frac{n\cos^{3}ny}{3n}\sin{ny}\right]_{-\pi}^{\pi}
    You have do this incorrectly. See here.
    Thanks from lawochekel
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  3. #3
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    Re: definite integral!

    Hello, lawochekel!

    You integrated incorrectly.

    We need the identity: . \cos^2\theta \:=\:\tfrac{1}{2}\left(1 + \cos2\theta\right)


    \int_{\text{-}\pi}^{\pi} \cos^2(ny)\,dy

    \int_{\text{-}\pi}^{\pi} \cos^2(ny)\,dy \;=\;\tfrac{1}{2}\int^{\pi}_{\text{-}\pi}\big[1 + \cos(2ny)\big]\,dy

    . . . . . . . . . . . =\; \tfrac{1}{2}\big[y + \tfrac{1}{2n}\sin(2ny)\bigg]^{\pi}_{\text{-}\pi}

    . . . . . . . . . . . =\;\tfrac{1}{2}\left[\pi + \tfrac{1}{2n}\sin(2\pi n)\right] - \tfrac{1}{2}\left[\text{-}\pi + \tfrac{1}{2n}\sin(\text{-}2\pi n)\right]

    . . . . . . . . . . . =\;\tfrac{1}{2}(\pi + 0) - \tfrac{1}{2}(\text{-}\pi + 0)

    . . . . . . . . . . . =\; \pi
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