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Homework Help: Integral of x^5secxdx

  1. Oct 3, 2011 #1
    Hi I m kinda stuck at this question.

    integral of x^5secxdx. I tried IBP n i couldnt carry on.
     
  2. jcsd
  3. Oct 3, 2011 #2
    Unfortunately, this integral is not expressible in terms of elementary functions.
     
  4. Oct 3, 2011 #3
    Ok my frene gave me the wrong info. int that interms of -1 to 1.
    I know the answer is 0 but why?
     
  5. Oct 3, 2011 #4
    Is the function [itex]f(x) = x^5 \sec{x}[/itex] even, odd or neither? When you have decided that, it should become clear as to why

    [itex]\int^{1}_{-1}{x^5 \sec{x}} dx = 0[/itex]
     
  6. Oct 3, 2011 #5
    I feel lost lol. How do i tell if it's even odd or neither?
     
  7. Oct 3, 2011 #6
    A function [itex]f[/itex] is even if [itex]f(x) = f(-x) \forall x \in domain[/itex]

    A function [itex]f[/itex] is odd if [itex]-f(x) = f(-x) \forall x \in domain[/itex]

    For your function, check [itex]f(x)[/itex] vs. [itex]f(-x)[/itex] vs. [itex]-f(x)[/itex].
     
    Last edited: Oct 3, 2011
  8. Oct 3, 2011 #7
    f(-x) = -x^5sec-x

    -f(x) = -(x^5secx)

    Is this the way?
     
  9. Oct 3, 2011 #8
    Sorry, I initially missed a negative sign in my last post.

    Yes, that's a way to check. And you know that sec(x) = 1/cos(x) and that cos(-x) = cos(x). Therefore, sec(-x) = sec(x). That implies that sec is an even function. However, (-x)^5 = -(x^5) implies that x^5 is an odd function. An even function times an odd function is itself an odd function.

    One property of an odd function, g(x), is that

    [itex]\int^{a}_{-a} g(x)dx = 0[/itex]
     
  10. Oct 3, 2011 #9
    On a related note, for an even function h(x),

    [itex]\int^{a}_{-a} h(x)dx = 2\int^{a}_{0} h(x)dx[/itex]

    If you graph some simple even and odd functions, you will see why this is the case.
     
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