# Solving ∫cosh^2(x)sinh(x)dx

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
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!

• SammyS
Mark44
Mentor

## 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.

epenguin
Homework Helper
Gold Member
sinh x dx is d(what) ?