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Can someone please help me with a quick integral?

  1. Aug 10, 2004 #1
    I don't know how to do fancy symbols, but

    Integral of 1/((sinh[x])^2) .dx

    I got as far as 1/2 Int(sech(2x)). d(2x), is this the right approach?

    Thanks .
  2. jcsd
  3. Aug 10, 2004 #2
    Your approach might work, but it seems to me like the integral could be easily handled by rewriting the hyperbolic sine in exponential form and then doing a substitution (or partial fractions (but that seems rather messy)).
    Last edited: Aug 10, 2004
  4. Aug 10, 2004 #3


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    this is probably easy if you remember the basic trig integrals and their analogies with hyperbolic trig integrals. i.e. remember the derivative of tan is sec^2, and sec = 1/cos.

    also the derivative of cot is -csc^2 and csc = 1/sin.

    Hence by analogy, we should try cosh/sinh as the antiderivative of 1/sinh. does it work?
  5. Aug 10, 2004 #4
    It write it in it’s exponent form then break it into two fractions and take each integral separately.
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