## anonymous 4 years ago 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

1. anonymous

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

2. anonymous

is it odd?

3. anonymous

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

4. anonymous

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

5. anonymous

heck no. tangent is odd, not even

6. anonymous

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

7. anonymous

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

8. anonymous

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!!!!