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How to find the integral of the secant function

  1. Jul 13, 2006 #1
    [tex] \int \sec hx[/tex]
    I solve it in this way:
    [tex] \int \arccos hx[/tex]
    [tex] \int \ln (x^2 + \sqrt{x^2 -1}) dx [/tex]

    Then, I substitute [tex] u = \ln (x + \sqrt{x^2 + 1})[/tex]
    then I get
    [tex] x\ln(x + \sqrt{x^2 + 1}) - \int x/\sqrt{x^2 + 1}dx[/tex]

    and then I substitute v = x^2 + 1

    [tex] x\ln(x + \sqrt{x^2 + 1}) -1/2 \int v^(1/2) dv [/tex]
    and finally I get the answer,
    [tex] x\ln(x + \sqrt{x^2 + 1}) -\sqrt{x^2 + 1} + C [/tex]

    Am I right?
     
    Last edited: Jul 13, 2006
  2. jcsd
  3. Jul 13, 2006 #2

    arildno

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    Do the Tanh(x/2) substitution.
     
  4. Jul 13, 2006 #3
    tan (x/2) substitution? How?
     
  5. Jul 13, 2006 #4

    arildno

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    Your second integral doesn't equal your first integral.
    sech(x) is NOT equal to arcosh(x)!!!!

    sech(x) is the RECIPROCAL* of cosh(x), arcosh(x) is the (functional) INVERSE of cosh(x).




    *Sometimes called the multiplicative inverse.
     
  6. Jul 13, 2006 #5

    StatusX

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    Do you know how to integrate sec(x)? Because a lot of times you can follow very similar procecures working with hyperbolic functions as with trigonometric ones. You just have to be careful about the signs of terms in certain identiies (eg, sin^2(x)+cos^2(x)=1 becomes cosh^2(x)-sinh^2(x)=1). Another approach would be to write sech out in terms of e^x and e^-x, and then use the substitution u=e^x.
     
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