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Thread: Integration by Parts

  1. July 17th 2006, 06:40 PM #1
    c_323_h
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    Integration by Parts

    Hi guys, need some help on these:

    1. \int x (lnx) dx

    2. \int x(cos5x)dx
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  2. July 17th 2006, 07:14 PM #2
    ThePerfectHacker
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    Quote Originally Posted by c_323_h
    Hi guys, need some help on these:

    1. \int x (lnx) dx
    Let,
    u=\ln x and, v'=x
    Thus,
    u'=1/x and, v=(1/2)x^2
    Thus,
    \frac{1}{2}x^2\ln x-\int \frac{1}{2} xdx
    Thus,
    \frac{1}{2}x^2\ln x-\frac{1}{4}x^2+C
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  3. July 17th 2006, 07:17 PM #3
    ThePerfectHacker
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    Quote Originally Posted by c_323_h

    2. \int x(cos5x)dx
    Let,
    u=x and v'=\cos 5x
    Thus,
    u'=1 and v=\frac{1}{5}\sin 5x
    Thus,
    \frac{1}{5}x\sin 5x-\int \frac{1}{5}\sin 5xdx
    Thus,
    \frac{1}{5}x\sin 5x+\frac{1}{25}\cos 5x+C

    (I am the quickest integrator in the west!)
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  4. July 17th 2006, 07:17 PM #4
    c_323_h
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    Quote Originally Posted by ThePerfectHacker
    Let,

    ...-\int \frac{1}{2} xdx
    why isn't it \frac{1}{2}x^2dx.
    Where did the x^2 dissappear to?
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  5. July 17th 2006, 07:22 PM #5
    ThePerfectHacker
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    Quote Originally Posted by c_323_h
    why isn't it \frac{1}{2}x^2dx.
    Where did the x^2 dissappear to?
    \int x^n=\frac{x^{n+1}}{n+1}, n\not =-1
    Thus,
    \int xdx=\frac{1}{2}x^2 but then you need to multiply by \frac{1}{2} thus,
    \frac{1}{2}\cdot \frac{1}{2}x^2=\frac{1}{4}x^2
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