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Homework Help: Solving ∫cosh^2(x)sinh(x)dx

  1. Aug 5, 2015 #1
    Solving this the simple way I got the right solution, but I also tried to solve it a different way and got it wrong. I want to know where I went wrong the second way.

    1. The problem statement, all variables and given/known data

    ∫cosh2(x)sinh(x)dx = ?

    2. Relevant equations
    cosh(x)sinh(x) = (1/2)*sinh(2x)

    3. The attempt at a solution
    The solution is simple by guessing and checking with the chain rule:
    ∫cosh2(x)sinh(x)dx = 1/3*cosh3(x)

    But then I try to manipulate the expression in the integral I get a different result:
    cosh2(x)sinh(x) = cosh(x)cosh(x)sinh(x) = cosh(x)*(1/2)*sinh(2x) = (ex+e-x)/2*(1/2)*(e2x-e-2x)/2 = (1/4)*((e3x-e-3x)/2+(ex-e-x)/2 = (1/4)*(sinh(3x)+sinh(x))
    Integrating that gives (1/4)*∫(sinh(3x)+sinh(x))dx = (1/4)*(∫sinh(3x)dx+∫sinh(x)dx) = (1/4)*((cosh(3x)/3)+cosh(x))
    I'm not sure how to directly compare the two results, but plugging in some number for x and solving each shows that they aren't equal. I'm quite sure the first solution is correct, so where did I go wrong in the second?
  2. jcsd
  3. Aug 5, 2015 #2


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    Staff: Mentor

    You can plug in numbers in between and see where it goes wrong.
    Do they differ by a constant only?
  4. Aug 5, 2015 #3


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    Gold Member

    I think they're the same.

    You can expand each answer out in terms of exponential functions.

    Easier than that, take the derivative of each and compare.
  5. Aug 5, 2015 #4
    You're right Sammy, they are the same. I expanded each answer and got (1/24)*(e3x+e-3x+3ex+3e-x) for both. I must have made some input mistake when I plugged in a number for x.

  6. Aug 6, 2015 #5


    Staff: Mentor

    A better strategy is to use an ordinary substitution: u = cosh(x), du = sinh(x)dx.
  7. Aug 7, 2015 #6


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    sinh x dx is d(what) ?
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