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Math Help - Integration by parts

  1. #1
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    Integration by parts

    [tex]\int x^2sin\pixdx[\math]

    why doesnt this latex thing work for me????

    i dont know how to integrate x^2sin(pi)xdx
    Last edited by jeph; January 29th 2007 at 06:16 PM.
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    Quote Originally Posted by jeph View Post
    [tex]\intx^2sin\pixdx[\math]

    why doesnt this latex thing work for me????
    Your "slash" for the closing bracket is in the wrong direction:
    \int x^2 sin \pi xdx

    Using spaces would be a good idea, too.

    Quote Originally Posted by jeph View Post
    i dont know how to integrate x^2sin(pi)xdx
    \int p dq = pq - \int q dp

    So
    \int x^2 sin(\pi x)dx
    Pick
    p = x^2 ==> dp = 2x dx
    dq = sin(\pi x) dx ==> q = -\frac{1}{\pi}cos(\pi x)

    Then
    \int x^2 sin(\pi x)dx = -\frac{1}{\pi} x^2 cos(\pi x) - \int (2x) \cdot -\frac{1}{\pi}cos(\pi x) dx  = -\frac{1}{\pi} x^2 cos(\pi x) + \frac{2}{\pi} \int x cos(\pi x) dx

    We can use the same process on the remaining integral:
    \int x cos(\pi x) dx = \frac{x}{\pi}sin(\pi x) - \int \frac{1}{\pi}sin(\pi x) dx  = \frac{x}{\pi}sin(\pi x) + \frac{1}{\pi ^2}cos(\pi x)

    Putting this result into our original integral:
    \int x^2 sin(\pi x)dx = -\frac{1}{\pi} x^2 cos(\pi x) + \frac{2}{\pi} \left ( \frac{x}{\pi}sin(\pi x) + \frac{1}{\pi ^2}cos(\pi x) \right )

     = -\frac{1}{\pi} x^2 cos(\pi x) + \frac{2x}{\pi ^2}sin(\pi x) + \frac{2}{\pi ^3}cos(\pi x)

    -Dan
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    nevermind, i got it. thanks!

    Quote Originally Posted by topsquark View Post
    Your "slash" for the closing bracket is in the wrong direction:
    \int x^2 sin \pi xdx

    Using spaces would be a good idea, too.


    \int p dq = pq - \int q dp

    So
    \int x^2 sin(\pi x)dx
    Pick
    p = x^2 ==> dp = 2x dx
    dq = sin(\pi x) dx ==> q = -\frac{1}{\pi}cos(\pi x)

    Then
    \int x^2 sin(\pi x)dx = -\frac{1}{\pi} x^2 cos(\pi x) - \int (2x) \cdot -\frac{1}{\pi}cos(\pi x) dx  = -\frac{1}{\pi} x^2 cos(\pi x) + \frac{2}{\pi} \int x cos(\pi x) dx

    We can use the same process on the remaining integral:
    \int x cos(\pi x) dx = \frac{x}{\pi}sin(\pi x) - \int \frac{1}{\pi}sin(\pi x) dx  = \frac{x}{\pi}sin(\pi x) + \frac{1}{\pi ^2}cos(\pi x)

    Putting this result into our original integral:
    \int x^2 sin(\pi x)dx = -\frac{1}{\pi} x^2 cos(\pi x) + \frac{2}{\pi} \left ( \frac{x}{\pi}sin(\pi x) + \frac{1}{\pi ^2}cos(\pi x) \right )

     = -\frac{1}{\pi} x^2 cos(\pi x) + \frac{2x}{\pi ^2}sin(\pi x) + \frac{2}{\pi ^3}cos(\pi x)

    -Dan
    how do you integrate the \int \frac{1}{\pi}sin(\pi x) dx into \frac{1}{\pi ^2}cos(\pi x)
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    Quote Originally Posted by jeph View Post
    nevermind, i got it. thanks!



    how do you integrate the \int \frac{1}{\pi}sin(\pi x) dx into \frac{1}{\pi ^2}cos(\pi x)
    That isn't quite what I did, you are off by a negative.

    In detail:
    \int \frac{1}{\pi} sin(\pi x) dx = \frac{1}{\pi} \int sin(\pi x) dx

    Let y = \pi x ==> dy = \pi dx

    So
    \int \frac{1}{\pi} sin(\pi x) dx = \frac{1}{\pi} \int \frac{dy}{\pi}sin(y) = \frac{1}{\pi ^2}(-cos(y)) = -\frac{1}{\pi ^2}cos(\pi x)

    -Dan
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    Quote Originally Posted by jeph View Post
    nevermind, i got it. thanks!



    how do you integrate the \int \frac{1}{\pi}sin(\pi x) dx into \frac{1}{\pi ^2}cos(\pi x)
    A useful rule to remember....

    Let F(x) be an antiderivative of f(x).

    Then, ( a\not = 0),
    \int f(ax+b)dx=\frac{1}{a}F(ax+b)+C

    Meaning whenever you have a linear function (of the form ax+b) you just take the coefficient with the "x" (the slope of line) and invert it.

    For example,
    \int (2x+1)^4 dx
    Since this is a linear expression inside,
    2x+1.
    You can use the rule I told you.
    Just find the anti-derivative the regular way,
    \int (2x+1)^4 dx = \frac{(2x+1)^5}{5}+C
    But of course this answer is wrong, you need to invert the slope.
    Thus,
    \frac{1}{2} \cdot \frac{(2x+1)^5}{5}+C

    Now let us find,
    \int \sin (\pi x) dx
    Note \pi x=\pi x+0 is a linear function in the special form we had above because \pi is just a number in front.
    Thus, to find the integral you just find the integral of sine which is -cosine and then invert the slope.
    Thus,
    \int \sin \pi x dx = -\frac{1}{\pi} \cos x+C.

    Learn, this rule, it saves much time.
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