# Solving ∫cosh^2(x)sinh(x)dx

1. Aug 5, 2015

### desquee

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. Aug 5, 2015

### Staff: Mentor

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

3. Aug 5, 2015

### SammyS

Staff Emeritus
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. Aug 5, 2015

### desquee

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!

5. Aug 6, 2015

### Staff: Mentor

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

6. Aug 7, 2015

### epenguin

sinh x dx is d(what) ?