Non-elementary Integral: Solving x^2sin(x)/(1+x^6) using Substitution Method

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Homework Statement



Solve the integral of x2sin(x) / (1+x6) from -pi/2 to pi/2.

Homework Equations



none

The Attempt at a Solution



Well I am supposed to do this using the substitution method, so I tried:

u = x2
du = 2x dx which doesn't cancel out any terms

u = 1+x6
du = 6x5 dx which again doesn't cancel anything out

u = sin(x)
du = cox(x) dx useless also

Is this possible with substitution? I seem to have tried every option for u. Is there another way to do this? (without taylor series and stuff like that)

Thanks for the help!Also, I tried doing this integral on the computer, and it said it cannot be solved because it is probably a "non-elementary integral", hence the title of the thread.
 
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The function is odd on that interval. f(-x)=-f(x).
 
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I had a similar question earlier. The result is a whole bunch of sine and cosine integrals and imaginary parts. I don't understand how one would do this, but go ahead and put it in the integrator (mathematica) and check the answer out. What level is this for?
 
For a symmetric function that is odd, the integral on the interval -a to a = 0 !
Thanks guys.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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