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Math Help - Integration by parts help

  1. #1
    Junior Member
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    Oct 2009
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    Integration by parts help

    Hi all,

    Can anybody show me how to get to this point I have tried a million different ways and don't know how it is done????

     <br />
\frac{\pi}{2} \int_{0}^\pi x(\pi-x)\sin(nx) dx = <br />
\frac{4}{n^3 \pi} [1-(-1)^n]<br />

    any help will be very greatly appreciated
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  2. #2
    MHF Contributor Amer's Avatar
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    Jordan
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    Quote Originally Posted by davefulton View Post
    Hi all,

    Can anybody show me how to get to this point I have tried a million different ways and don't know how it is done????

     <br />
\frac{\pi}{2} \int_{0}^\pi x(\pi-x)\sin(nx) dx = <br />
\frac{4}{n^3 \pi} [1-(-1)^n]<br />

    any help will be very greatly appreciated
    \int x(\pi -x)\sin (nx)\cdot dx

    let v=x(\pi -x ) and du =sin(nx) , dv=(\pi -x ) -x .. u =-cos(nx)/n

    \int x(\pi -x)\sin (nx)\cdot dx =x(\pi -x )\left(\frac{-\cos nx}{n}\right) + \int ((\pi -x ) -x)\frac{\cos nx}{n} \cdot dx

    \int x(\pi -x)\sin (nx)\cdot dx =x(\pi -x )\left(\frac{-\cos nx}{n}\right) + \int (\pi-x)\frac{\cos nx}{n} \cdot dx - \int x \frac{\cos nx}{n} \cdot dx

    the two integrals have the same idea by parts

    let v=(\pi -x) and du=cos(nx) ...dv=-1 u=sin(nx)/n in the first integral
    let v=x and du=cos(nx) ....dv=1 u=sin(nx)/n in the second integral
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