Integrating e^-2x.tanh(x) - Solving for the Antiderivative

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In summary, the homework statement is to find the integral \int e^{-2x}\tanh x\,dx. The attempt at a solution is to integrate by parts, but gets stuck. The answer is to integrate by parts then let u = e-x or maybe let u = 1+e-2x. The final answer is \frac{e^{-2x}}{2}-\ln (1+e^{-2x}) but the answer in my copybook is \frac{e^{-2x}}{2}-\ln (1+e^{-2x})
  • #1
DryRun
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Homework Statement

Find [tex]\int e^{-2x}\tanh x\,dx[/tex]

The attempt at a solution

[tex]\int e^{-2x}\tanh x\,dx
\\=\int e^{-2x}\times \frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx
\\=\int \frac{e^{-x}-e^{-3x}}{e^x+e^{-x}}\,dx
[/tex]

Then i have no idea how to proceed.
 
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  • #2
Hint: can you do anything with partial fractions?
 
  • #3
I've already tried to split it:
[tex]\int \frac{e^{-x}}{e^x+e^{-x}}-\int \frac{e^{-3x}}{e^x+e^{-x}}[/tex]But got stuck again.

The answer is: [tex]\frac{e^{-2x}}{2}-\ln (1+e^{2x})[/tex]

I'm thinking maybe integration by parts?
 
  • #4
sharks said:
I've already tried to split it:
[tex]\int \frac{e^{-x}}{e^x+e^{-x}}-\int \frac{e^{-3x}}{e^x+e^{-x}}[/tex]But got stuck again.

The answer is: [tex]\frac{e^{-2x}}{2}-\ln (1+e^{2x})[/tex]

I'm thinking maybe integration by parts?
[itex]\displaystyle \frac{e^{-x}}{e^x+e^{-x}}=\frac{e^{-2x}}{1+e^{-2x}}[/itex]

[itex]\displaystyle \frac{e^{-3x}}{e^x+e^{-x}}=\frac{e^{-4x}}{1+e^{-2x}}\ .[/itex] Then let u = e-x or maybe let u = 1+e-2x .
 
  • #5
[tex]\int \frac{e^{-2x}}{1+e^{-2x}}\,.dx-\int \frac{e^{-4x}}{1+e^{-2x}}\,.dx[/tex]
Let [itex]u =e^{-x}
\\Then \, \frac{du}{dx}=-e^{-x}=-u[/itex]
[tex]=\int \frac{-u}{1+u^2}\,.du-\int \frac{-u^3}{1+u^2}\,.du[/tex]
[tex]=-\frac{1}{2}\ln (1+u^2) + \int u\,.du - \int \frac{u}{(1+u^2)}\,.du[/tex]
[tex]=-\frac{1}{2}\ln (1+u^2) + \frac{u^2}{2} - \frac{1}{2}\ln (1+u^2)[/tex]
[tex]=\frac{u^2}{2}-\ln (1+u^2)[/tex]
Substituting for u, we have the final answer:
[tex]=\frac{e^{-2x}}{2}-\ln (1+e^{-2x})[/tex]
But the answer in my copybook is:
[tex]=\frac{e^{-2x}}{2}-\ln (1+e^{2x})[/tex]
By the way, isn't there a need to add the arbitrary constant of integration?
 
Last edited:
  • #6
Your result is correct, and add a constant.

ehild
 
  • #7
Hi ehild!

Thank you for your confirmation. :smile:
 

Related to Integrating e^-2x.tanh(x) - Solving for the Antiderivative

What is the function e^{-2x}.tanh(x)?

The function e^{-2x}.tanh(x) is a combination of two mathematical functions: the exponential function e^{-2x} and the hyperbolic tangent function tanh(x). This function is frequently used in physics and engineering to model various phenomena.

What is the domain and range of e^{-2x}.tanh(x)?

The domain of e^{-2x}.tanh(x) is all real numbers, as both the exponential and hyperbolic tangent functions are defined for all values of x. The range of this function is from -1 to 1, as the hyperbolic tangent function has a maximum value of 1 and a minimum value of -1.

How do you integrate e^{-2x}.tanh(x)?

The integral of e^{-2x}.tanh(x) can be found using the substitution method. Let u = tanh(x), then du = sech^2(x) dx. Substituting into the integral gives the new integral of e^{-2x} u du. This can then be solved using integration by parts or the tabular method.

What is the significance of e^{-2x}.tanh(x) in science?

E^{-2x}.tanh(x) has various applications in physics and engineering. It can be used to model the behavior of damped harmonic oscillators, heat transfer in materials, and the dynamics of electrical circuits. It also has applications in statistics and probability, such as in the study of Brownian motion and random walks.

How does the graph of e^{-2x}.tanh(x) look like?

The graph of e^{-2x}.tanh(x) has an S-shaped curve, similar to the hyperbolic tangent function. It approaches 0 as x tends to negative infinity and approaches 1 as x tends to positive infinity. It also has an inflection point at (0,0) where the slope changes from positive to negative.

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