What is the error in my approach to solving ∫cosh^2(x)sinh(x)dx?

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Homework Help Overview

The discussion revolves around the integral ∫cosh²(x)sinh(x)dx, with participants exploring different approaches to solving it. The subject area includes hyperbolic functions and integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve the integral using two different methods, one involving direct integration and the other manipulating the integrand. They express confusion over the discrepancy between the results obtained from each method.
  • Some participants suggest checking if the two results differ by a constant and recommend comparing derivatives of the results to clarify the situation.
  • One participant proposes using a substitution method as an alternative approach.

Discussion Status

The discussion is active, with participants exploring the validity of the original poster's results. There is a recognition that both approaches may yield equivalent results, and some participants have confirmed this through expansion and derivative comparison.

Contextual Notes

There is mention of potential input mistakes when evaluating the results numerically, indicating that participants are working under the constraints of ensuring accuracy in their calculations.

desquee
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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?
 
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You can plug in numbers in between and see where it goes wrong.
Do they differ by a constant only?
 
desquee said:
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.
 
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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desquee said:

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.
 
sinh x dx is d(what) ?
 

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