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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

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  1. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    is it odd?

  3. anonymous
    • 4 years ago
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    I talked to a friend that said they logically deduced because the top is odd and the bottom is even...

  4. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    heck no. tangent is odd, not even

  6. anonymous
    • 4 years ago
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    denominator is even, |x| is even, but tangent is odd, so the whole thing is odd

  7. anonymous
    • 4 years ago
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    The function is odd as everything except for tanx is even.

  8. anonymous
    • 4 years ago
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    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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