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Math Help - [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.

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

    from there, I find:

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

    and get:

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

    and then the answer I get is:

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

    But the book gives this as the answer:
    \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 ...

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

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

    \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 ...

    u = x
    du = dx

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

    \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 =

     -\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.

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

    I set:

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

    and get:

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

    and from there I integrate by partial fractions:

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

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

    then I get:

    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?
    \int\frac{2x}{2x+1}dx=\int\underset{1}{\underbrace  {\frac{2x+1}{2x+1}}}dx-\int\frac{1}{2x+1}dx

    .........._ =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

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