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Thread: [SOLVED] Issue with a Integration by parts problem

  1. #1
    Junior Member Coco87's Avatar
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    [SOLVED] Issue with a Integration by parts problem

    Hey,
    I've run across a problem here that I can't seem to figure out. I need to Integrate a problem using the Integration by parts method.

    $\displaystyle \int x^2\sin{(\pi x)} dx$

    from there, I find:

    $\displaystyle u=x^2$
    $\displaystyle du=2x$
    $\displaystyle dv=\sin{(\pi x)}dx$
    $\displaystyle v=\frac{-1}{\pi}\cos{(\pi x)}$

    and get:

    $\displaystyle \frac{-1}{\pi}x^2 \cos{(\pi x)}-\int \frac{-2x}{\pi}\cos{(\pi x)}dx$

    and then the answer I get is:

    $\displaystyle \frac{-1}{\pi}x^2\cos{(\pi x)}+\frac{2x}{\pi^2}\sin{(\pi x)} + C$

    But the book gives this as the answer:
    $\displaystyle \frac{-1}{\pi}x^2\cos{(\pi x)}+\frac{2x}{\pi^2}\sin{(\pi x)}+\frac{2}{\pi^3}\cos{(\pi x)}+C$

    I'm completely stumped as to where they got the last part (before the constant C) from.

    Anyone?
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    you have to use parts twice ...

    $\displaystyle u = x^2$
    $\displaystyle du = 2x \, dx$

    $\displaystyle dv = \sin(\pi x) \, dx$
    $\displaystyle v = -\frac{1}{\pi} \cos(\pi x)$

    $\displaystyle \int x^2 \sin(\pi x) \, dx = -\frac{x^2}{\pi} \cos(\pi x) + \frac{2}{\pi} \int x \cos(\pi x) \, dx$

    using parts a second time for the last integral expression ...

    $\displaystyle u = x$
    $\displaystyle du = dx$

    $\displaystyle dv = \cos(\pi x) \, dx$
    $\displaystyle v = \frac{1}{\pi} \sin(\pi x)$

    $\displaystyle \int x^2 \sin(\pi x) \, dx = -\frac{x^2}{\pi} \cos(\pi x) + \frac{2x}{\pi^2} \sin(\pi x) - \frac{2}{\pi^2} \int \sin( \pi x) \, dx =$

    $\displaystyle -\frac{x^2}{\pi} \cos(\pi x) + \frac{2x}{\pi^2} \sin(\pi x) + \frac{2}{\pi^3} \cos(\pi x) + C$
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  3. #3
    Junior Member Coco87's Avatar
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    Ah, I didn't know you could build it like that.

    Thanks!
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  4. #4
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Coco87 View Post
    Sorry to bring this thread back up, but I figured this would be better than starting a new one. I have run into another brick wall that is driving me insane (see following picture: )

    It seems simple, yet I'm not getting the answer in the book, but rather I'm looping through the Integration By Parts method over and over again.

    $\displaystyle \int{\ln{(2x+1)}dx}$

    I set:

    $\displaystyle u=\ln{(2x+1)}$ $\displaystyle dv=dx$
    $\displaystyle du=\frac{2}{2x+1}$ $\displaystyle v=x$

    and get:

    $\displaystyle x\ln{(2x+1)}-\int{\frac{2x}{2x+1}}$

    and from there I integrate by partial fractions:

    $\displaystyle 2\int{x\frac{1}{2x+1}}$

    $\displaystyle u=\frac{1}{2x+1}$ $\displaystyle dv=x$
    $\displaystyle du=\frac{2}{(2x+1)^2}$ $\displaystyle v=\frac{x^2}{2}$

    then I get:

    $\displaystyle 2[\frac{x^2}{2(2x+1)}-\int{\frac{x^2}{(2x+1)^2}}]$

    I might be wrong, but it seems like and endless loop to me

    Could anyone help me out?
    $\displaystyle \int\frac{2x}{2x+1}dx=\int\underset{1}{\underbrace {\frac{2x+1}{2x+1}}}dx-\int\frac{1}{2x+1}dx$

    .........._$\displaystyle =x-\frac{1}{2}\ln|2x+1|$
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by bkarpuz View Post

    .........._$\displaystyle =x-\frac{1}{2}\ln|2x+1|$
    $\displaystyle +k.$
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  6. #6
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle +k.$
    sure!
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