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So I have a two fold question, one I believe is simple but my algebra seems to be off, the other involves the trapezoidal rule of integration using Mathematica as an aid. Here they are:

$1.\quad \displaystyle \int_{-\infty}^{\infty} \frac{\operatorname{sech}(x)}{x^2+1} dx = \int_{-1}^{1} \operatorname{sech}\left(\frac{1}{1-t^2}\right)\frac{t^2+1}{t^4-t^2+1} dt$

I know I need to let $x = \frac{t}{1-t^2}$ and take the limits as $t \to \infty$,change the limits of integration and do the same for $t \to -\infty$ but I can't seem to nail it down. Why are my limits going to be $-1$ and $1$?

$2$. Space five points equally from $-1$ to $1$ and compute the four trapezoid approximation of $\int_{-1}^{1} \mathrm{sech}(\frac{1}{1-t^2})\frac{t^2+1}{t^4-t^2+1} dt$ using Mathematica to evaluate $\operatorname{sech}(x)$. To be honest, I'm not really sure what the question is asking. Am I breaking the integral up into four integrations the first of which is from $-1$ to $-0.5$? How do I use Mathematica to evaluate? Any help is appreciated.

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    "Am I breaking the integral up into four integrations..." - yep, you interpreted correctly. Try it out!2012-02-18
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    For the first question: what values of $t$ will make $\dfrac{t}{1-t^2}$ take the values $-\infty$ and $+\infty$?2012-02-18
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    @J.M. Thank you. I entered the integral from $-1$ to $-0.5$ into Mathematica but it gives me back an answer which still involves sech (using the original function).2012-02-18
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    the second question was asking you to use the trapezoidal rule over the four separate panels you made out of the interval $(-1,1)$ for the evaluation, and not an analytical evaluation... :)2012-02-18
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    LOL. I didn't clarify. This is for one of my tutoring students and the professor specifically requests that the students use Wolfram Alpha or Mathematica to evaluate sech(x).2012-02-18
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    Ah, yes. You indeed have to evaluate the hyperbolic secant to be able to use the trapezoidal rule. But, first things first: I assume you at least know how to evaluate $\lim\limits_{u\to\infty}\mathrm{sech}(x)$? (Since $\mathrm{sech}(x)$ is even, that would also be the value of the limit as $x\to\-\infty$.)2012-02-18
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    Yes, I get how to do it by hand and sketched a graph to help my student visually. In fact, the computing is the hardest part of this class since I have never taken or tutored a Calculus class that requires you using it.2012-02-18
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    If you're ubstituting $$\frac{t}{{1 - {t^2}}} = x$$ why do you get $$\operatorname{sech} \left( {\frac{1}{{1 - {t^2}}}} \right)$$ instead of $${\operatorname{sech} \left( {\frac{t}{{1 - {t^2}}}} \right)}$$?2012-02-18

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