Solving ∫cosh^2(x)sinh(x)dx

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  • #1
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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. Homework Statement

∫cosh2(x)sinh(x)dx = ?

Homework Equations


cosh(x)sinh(x) = (1/2)*sinh(2x)

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?
 

Answers and Replies

  • #2
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You can plug in numbers in between and see where it goes wrong.
Do they differ by a constant only?
 
  • #3
SammyS
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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. Homework Statement

∫cosh2(x)sinh(x)dx = ?

Homework Equations


cosh(x)sinh(x) = (1/2)*sinh(2x)

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?
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.
 
  • #4
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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.

Thanks!
 
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  • #5
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The Attempt at a Solution


The solution is simple by guessing and checking with the chain rule
A better strategy is to use an ordinary substitution: u = cosh(x), du = sinh(x)dx.
 
  • #6
epenguin
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sinh x dx is d(what) ?
 

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