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Thread: definite integral!

  1. #1
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    definite integral!

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

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

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

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

    $\displaystyle 2\left[0\right] $

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

    thanks!
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  2. #2
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    Re: definite integral!

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

    $\displaystyle \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: .$\displaystyle \cos^2\theta \:=\:\tfrac{1}{2}\left(1 + \cos2\theta\right)$


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

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

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

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

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

    . . . . . . . . . . . $\displaystyle =\; \pi$
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