Integrate x^5secxdx to Solving the Tricky Integration Problem

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islubio
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Hi I m kinda stuck at this question.

integral of x^5secxdx. I tried IBP n i couldn't carry on.
 
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Unfortunately, this integral is not expressible in terms of elementary functions.
 
Ok my frene gave me the wrong info. int that interms of -1 to 1.
I know the answer is 0 but why?
 
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]
 
I feel lost lol. How do i tell if it's even odd or neither?
 
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:
f(-x) = -x^5sec-x

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

Is this the way?
 
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]
 
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.