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How do you integrate sec^3(x)

  1. Jan 19, 2009 #1
    This is not homework, but I'm just wondering, how do you integrate this deceptive looking integrand to get what Wolfram has?

    I don't get why the answer has an inverse hyperbolic function. Please teach me!
  2. jcsd
  3. Jan 19, 2009 #2


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    [tex]\int sec^3xdx = \int secx (sec^2x dx)[/tex]

    Integration by parts and then use the identity [itex]sec^2x=tan^2x+1[/itex]
  4. Jan 19, 2009 #3
    That method leads to (1/2)sec(x)tan(x) + (1/2)ln(sec(x) + tan(x)) + C. I am interested in getting an inverse hyperbolic function as displayed on Wolfram.

    I do not know how inverse hyperbolic functions are related to integrals. The only success I've had was integrating sec(x) into 2tanh^-1(tan(x/2))
  5. Jan 19, 2009 #4
    Maybe you should post what wolfram got?
  6. Jan 19, 2009 #5
  7. Jan 20, 2009 #6


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    Inverse hyperbolic functions can be written in terms of logarithms. In particular,

    [tex]\operatorname{artanh}(x) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)[/tex]

    So playing around with your logarithm you can probably get the artanh function they give out. (You may need to add a constant to your result to get to theirs).
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