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

  • Thread starter desquee
  • Start date
In summary, when trying to solve the homework equation cosh2(x)sinh(x)dx, one should use the substitution u = cosh(x), du = sinh(x)dx.sinh x dx is d(what). If one makes a mistake when plugging in a number for x, the results may not be correct.
  • #1
desquee
18
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. 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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  • #2
You can plug in numbers in between and see where it goes wrong.
Do they differ by a constant only?
 
  • #3
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.
 
  • #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.

Thanks!
 
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  • #5
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.
 
  • #6
sinh x dx is d(what) ?
 

What is the integration of cosh^2(x)sinh(x)?

The integration of cosh^2(x)sinh(x) is given by (1/3)cosh^3(x) + C.

How do I solve the integral of cosh^2(x)sinh(x)?

To solve the integral of cosh^2(x)sinh(x), you can use the substitution method or integration by parts. Alternatively, you can use a trigonometric identity to rewrite cosh^2(x)sinh(x) as a polynomial and then integrate.

What is the best method to solve ∫cosh^2(x)sinh(x)dx?

The best method to solve ∫cosh^2(x)sinh(x)dx depends on your familiarity and comfort with different integration techniques. It is recommended to try different methods and see which one works best for you.

Is there a shortcut to solve ∫cosh^2(x)sinh(x)dx?

There is no specific shortcut to solve ∫cosh^2(x)sinh(x)dx. However, you can use certain trigonometric identities or integration by parts to simplify the integral and make it easier to solve.

What are some tips for solving ∫cosh^2(x)sinh(x)dx?

Some tips for solving ∫cosh^2(x)sinh(x)dx include using trigonometric identities, trying different integration techniques, and carefully checking your work for any mistakes. It is also helpful to practice and familiarize yourself with different integration methods.

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