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Math Help - Definite integral

  1. #1
    MHF Contributor alexmahone's Avatar
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    Definite integral

    Find

    \frac{1}{\pi}\left[\int_{-\pi}^0(\pi+t)\sin ntdt+\int_0^\pi(\pi-t)\sin ntdt\right]

    for n=1, 2, 3,\ldots.

    My working:


    =\frac{1}{\pi}\left[\int_{-\pi}^0(\pi+t)\sin ntdt+\int_0^\pi(\pi-t)\sin ntdt\right]

    =\frac{1}{\pi}\left[\left.\left(-\frac{\pi}{n}\cos nt\right)\right|_{-\pi}^0+\int_{-\pi}^0 t\sin ntdt+\left.\left(-\frac{\pi}{n}\cos nt\right)\right|_0^\pi-\int_0^\pi t\sin ntdt\right]

    =\frac{1}{\pi}\left[-\frac{\pi}{n}\{1-(-1)^n\}+\int_{-\pi}^0 t\sin ntdt-\frac{\pi}{n}\{(-1)^n-1\}-\int_0^\pi t\sin ntdt\right]

    =\frac{1}{\pi}\left[\int_{-\pi}^0 t\sin ntdt-\int_0^\pi t\sin ntdt\right]

    =\frac{1}{n^2\pi}\left[\int_{-n\pi}^0u\sin udu-\int_0^{n\pi} u\sin udu\right]\hspace{5 mm}\left(u=nt, t=\frac{u}{n}\right)

    =\frac{1}{n^2\pi}[\left.(\sin u-u\cos u)\right|_{-n\pi}^0-\left.(\sin u-u\cos u)\right|_0^{n\pi}]

    =\frac{1}{n^2\pi}[n\pi(-1)^n+n\pi(-1)^n]

    =\frac{2(-1)^n}{n}

    However, the answer given in the book is 0. Where have I gone wrong?
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Definite integral

    Quote Originally Posted by alexmahone View Post
    =\frac{1}{n^2\pi}[\left.(\sin u-u\cos u)\right|_{-n\pi}^0-\left.(\sin u-u\cos u)\right|_0^{n\pi}]

    =\frac{1}{n^2\pi}[n\pi(-1)^n+n\pi(-1)^n]
    You dropped a minus sign in the first term.
    It should be:

    0-(-(-n\pi)\cos(-n\pi)) = -n\pi(-1)^n
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  3. #3
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    Re: Definite integral

    I suggest that, instead of actually computing the integral, you exploit the symmetries. That's much faster and less error-prone:

    \frac{1}{\pi} \left[ \int_{-\pi}^0 (\pi+t) \sin(nt) dt + \int_0^\pi (\pi-t) \sin(nt) dt \right]
    = \int_{-\pi}^\pi \sin(nt) dt + \frac{1}{\pi} \left[ \int_{-\pi}^0 t \sin(nt) dt - \int_0^\pi t \sin(nt) dt \right]
    = 0 + \frac{1}{\pi} \left[ \int_0^\pi t \sin(nt) dt - \int_0^\pi t \sin(nt) dt \right]
    = 0
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