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Evaluate the definite integral

  1. Oct 21, 2006 #1

    I've been having difficulty with this integral for some time now and any help would be gratly appreciated.

    [tex]\int\frac{x^2 \sin x}{1+x^6}dx[/tex]

    this is a definite integral from -pi/2 to pi/2

    The sinx has been giving me problems because if I set u = to any part of the equation I can't write sin(u)

    for example

    u = x^2 du/2 = xdx
    u = 1+x^6 du/6 = x^5dx
    u = 1+x^3 du/2 = x^2dx

    in all these cases I still get stuck with the sinx..

    hints on how to approach this equation would be ideal becasue I need to learn how to do this myself.

    Thank you!!
  2. jcsd
  3. Oct 21, 2006 #2


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    Homework Helper

    You won't be able to get the indefinite integral for that thing. But fortunately, you can use the fact the function is odd and the integration region is even.
  4. Oct 21, 2006 #3
    hey thanks for the help everyone.

    StatusX, what you're saying is that I can just write the final result as

    [tex]\int \frac{x^2 \sin x}{1+x^6}dx = 0 [/tex]

    and the function is odd because of sin(x) right?

    Last edited: Oct 21, 2006
  5. Oct 21, 2006 #4


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    Homework Helper

    Yea, because it is an even function times an odd function.
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