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Math Help - Integration by Parts

  1. #1
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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. #2
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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. #3
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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. #4
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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. #5
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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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