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Math Help - Trig Integral - Direct Comparison Theorem

  1. #1
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    Trig Integral - Direct Comparison Theorem

    Use the Comparison Theorem to determine whether the integral is convergent or divergent.

    I can easily see that:

    \frac{1}{x \cos{x}} \leq \frac{1}{x} on 0 to \frac{\pi}{2}

    But I also know that \int_0^{\frac{\pi}{2}}\frac{1}{x}dx is divergent because \ln{0} = -\infty

    So, that substitution doesn't tell me anything about my original integral. Can anyone suggest another substitution that might help me?

    Thanks.
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  2. #2
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    Re: Trig Integral - Direct Comparison Theorem

    Quote Originally Posted by joatmon View Post
    Use the Comparison Theorem to determine whether the integral is convergent or divergent.

    I can easily see that:

    \frac{1}{x \cos{x}} \leq \frac{1}{x} on 0 to \frac{\pi}{2} is that so?
    ...
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  3. #3
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    Re: Trig Integral - Direct Comparison Theorem

    Ack! I was thinking of x>=1. Thanks for the graph.
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