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I Integrating sqrt(1+x^2)

  1. Jan 23, 2017 #1
    So I am trying compute ##\displaystyle \int \sqrt{1+x^2}dx##. To start, I make the substitution ##u=\tan x##. After manipulation, this gives us ##\displaystyle \int |\sec u| \sec^2u ~du##. How do I get rid of the absolute value sign, so that I can go about integrating ##\sec^3 u##? Is there an argument that shows that ##\sec u## is always positive or always negative?
     
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  3. Jan 23, 2017 #2

    Svein

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    Try substituting [itex]x=\sinh(u) [/itex]...
     
  4. Jan 23, 2017 #3

    stevendaryl

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    Just a suggestion: Instead of using [itex]x = tan(u)[/itex], you might be better off with [itex]x = sinh(u)[/itex].

    As to your original question, [itex]sec(u) = \frac{1}{cos(u)}[/itex]. So [itex]sec(u) > 0[/itex] whenever [itex]cos(u) > 0[/itex], which means for [itex]|u| < \frac{\pi}{2}[/itex]. But from your substitution, [itex]x = tan(u)[/itex], there is no reason to consider [itex]|u| > \frac{\pi}{2}[/itex], because the range [itex]-\infty < x < +\infty[/itex] maps to the range [itex]-\frac{\pi}{2} < u < +\frac{\pi}{2}[/itex].
     
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