anonymous
  • 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
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
is it odd?
anonymous
  • 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
  • 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
  • anonymous
heck no. tangent is odd, not even
anonymous
  • anonymous
denominator is even, |x| is even, but tangent is odd, so the whole thing is odd
anonymous
  • anonymous
The function is odd as everything except for tanx is even.
anonymous
  • 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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