So, I had this question on my Calculus 2 test a little bit ago and I could not for the life of me figure out how to solve this. integrate (absolute value of x tanx)/(1 - 3x^2 + x^8) from -.5 to .5

Question

So, I had this question on my Calculus 2 test a little bit ago and I could not for the life of me figure out how to solve this.  integrate (absolute value of x tanx)/(1 - 3x^2 + x^8) from -.5 to .5

anonymous · Answer

Actually I had to prove that it equals zero.
$$\int\limits_{.5}^{-.5}(\left| x \right|tanx)/(1-3x^2 +x^8)=0$$

anonymous · Answer

is it odd?

anonymous · Answer

I talked to a friend that said they logically deduced because the top is odd and the bottom is even...

anonymous · Answer

that is, for the integrand, is 
$$f(-x)=f(x)$$? if so, then the integral has to equal zero because 
$$\int_{-a}^af(x)dx=0$$ if f is odd

anonymous · Answer

heck no.  tangent is odd, not even

anonymous · Answer

denominator is even, |x| is even, but tangent is odd, so the whole thing is odd

anonymous · Answer

The function is odd as everything except for tanx is even.

anonymous · Answer

Ahhhh, so all I had to do was to see if it was odd?  Goodness, I was trying to some how separate it the denominator to get tan^-1.  Thank you for your help!!!!