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Integrate (x^2 / sinx) / (1 + x^6)

  1. Dec 27, 2005 #1
    Integrate (x^2 / sinx) / (1 + x^6):

    How to do this? I just need some clues...
     
  2. jcsd
  3. Dec 27, 2005 #2

    siddharth

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    Do you mean
    [tex] \int \frac{x^2 dx}{(\sin x)(1+x^6)} [/tex]?

    Also, are you sure that it is not a definite integral?
     
  4. Dec 27, 2005 #3
    Sorry...should be (x^2 )( sinx) / (1 + x^6):
     
  5. Dec 27, 2005 #4
    and from x = - pi / 2 to x = pi / 2
     
  6. Dec 27, 2005 #5

    siddharth

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    Ah! That makes the problem solvable.

    Here is a hint:Find if your function is even or odd
     
  7. Dec 27, 2005 #6
    Erm...may I know what is odd and even function?
     
  8. Dec 27, 2005 #7
    A function, f(x), is even if

    f(-x) = f(x)

    and odd if

    f(-x) = -f(x)

    the integral of an odd from function from -a to a is zero. and the integral of an even function from -a to a is equal to two times the integral of that function from 0 to a.
     
  9. Dec 27, 2005 #8
    hint2: sin(x) = - sin(-x)
     
  10. Dec 28, 2005 #9
    Let x=(tan@)^3
     
  11. Dec 28, 2005 #10
    What`s the point to let x=(tan@)^3 ?
     
  12. Dec 28, 2005 #11
    Now that i think about it, such substitution might not work...


    Do you have a limit for your integral? Are you testing for convergence?
     
    Last edited: Dec 28, 2005
  13. Dec 28, 2005 #12

    arildno

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    Use the even or odd suggestion. VISUALIZE what it means that a function is odd, and how that will affect the value of a definite integral over that function between limits -a and a.
     
  14. Dec 28, 2005 #13
    By plotting the graph of sinx , the area within [tex]\frac{-\Pi}{2}[/tex] and [tex]\frac{\Pi}{2}[/tex] is zero and also know that the function is an odd function. So, what I get is only the value of sinx , how about [tex]\frac{x^2}{1+x^6}[/tex]?
     
  15. Dec 28, 2005 #14

    arildno

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    Is the function [itex]f(x)=\frac{x^{2}\sin(x)}{1+x^{6}}[/itex] even or odd?

    The answer to that is SUFFICIENT to answer what your definite integral is going to be.
     
  16. Dec 28, 2005 #15
    Ya. Thanks. Finally i catch the ball. But how should I answer this question in a proper way? Is there any other answering method for this question?
     
  17. Dec 28, 2005 #16

    arildno

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    It IS a proper way!!
     
  18. Dec 28, 2005 #17

    VietDao29

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    There's a little proof:
    #1. If f(x) is an odd function then: [tex]\int \limits_{-\alpha} ^ \alpha f(x) dx = 0[/tex].
    #2. If f(x) is an even function then: [tex]\int \limits_{-\alpha} ^ \alpha f(x) dx = 2 \ \int \limits_{0} ^ \alpha f(x) dx = 2 \ \int \limits_{-\alpha} ^ 0 f(x) dx[/tex].
    You can easily see this from a graph.
    I'll do #2, leaving #1 for you.
    [tex]\int \limits_{-\alpha} ^ \alpha f(x) dx = \int \limits_{-\alpha} ^ 0 f(x) dx + \int \limits_{0} ^ \alpha f(x) dx[/tex]
    Let's first integrate this:
    [tex]\int \limits_{-\alpha} ^ 0 f(x) dx[/tex]. Substitude x = -u, so: dx = -du. x = 0 => u = 0 [itex]x = -\alpha \Rightarrow u = \alpha[/itex]
    [tex]\int \limits_{-\alpha} ^ 0 f(x) dx = - \int \limits_{\alpha} ^ 0 f(-u) du = \int \limits_{0} ^ \alpha f(-u) du[/tex]. Using the fact that f(-u) = f(u), you'll have:
    [tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ \alpha f(x) dx[/tex], u is the dummy variable.
    So:
    [tex]\int \limits_{-\alpha} ^ \alpha f(x) dx = \int \limits_{-\alpha} ^ 0 f(x) dx + \int \limits_{0} ^ \alpha f(x) dx = \int \limits_{0} ^ \alpha f(x) dx + \int \limits_{0} ^ \alpha f(x) dx = 2 \ \int \limits_{0} ^ \alpha f(x) dx[/tex]. (Q.E.D)
    Can you do the same to #1, and the other part of #2?
     
  19. Dec 28, 2005 #18
    Thanks. The explanation is very clear. :)
     
  20. Dec 28, 2005 #19

    This surely is convergent. Let u= x^3
     
  21. Dec 28, 2005 #20
    how come [tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ \alpha f(x) dx[/tex]

    Shouldn`t it be [tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ {-\alpha} f(x) dx[/tex]
     
    Last edited: Dec 28, 2005
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