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- anonymous

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

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- anonymous

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- anonymous

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

- anonymous

is it odd?

- anonymous

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

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- 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

- anonymous

heck no. tangent is odd, not even

- anonymous

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

- anonymous

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

- 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!!!!

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