How Do You Solve the Integral of (1-e^-x)^(1/2)?

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Homework Help Overview

The problem involves finding the integral of the expression (1-e^-x)^(1/2). This falls under the subject area of calculus, specifically focusing on integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster seeks assistance with the integral, while one participant suggests a substitution method involving t^2=1-e^(-x). Other participants share their own expressions for the integral and discuss the complexity of the problem in the context of an introductory physics class.

Discussion Status

Participants are actively sharing their attempts and reasoning, with some providing specific expressions they derived. There is recognition of the problem's difficulty, and suggestions for alternative approaches are being explored.

Contextual Notes

Some participants express that the problem may be more suitable for a higher-level calculus course rather than an introductory physics class, indicating a potential mismatch in expected difficulty.

osy044
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Homework Statement


what is the integral of (1-e^-x)^(1/2)?
HELP~~


Homework Equations





The Attempt at a Solution

 
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What do you think about the substitution t^2=1-e^(-x)= ?

ehild
 
I think it is
[tex] \frac{\sqrt{1-e^{-x}} \left(e^{x/2} x-2 \sqrt{e^x-1}+2 e^{x/2} \log<br /> \left(-e^{-x/2} \sqrt{e^x-1}-1\right)\right)}{\sqrt{e^x-1}}[/tex]

Is that what you got as well?

By the way, I find this a rather hard calculus problem for an introductory physics class.
 
CompuChip said:
I think it is
[tex] \frac{\sqrt{1-e^{-x}} \left(e^{x/2} x-2 \sqrt{e^x-1}+2 e^{x/2} \log<br /> \left(-e^{-x/2} \sqrt{e^x-1}-1\right)\right)}{\sqrt{e^x-1}}[/tex]

Is that what you got as well?

By the way, I find this a rather hard calculus problem for an introductory physics class.

I was too lazy to check either formula. I left e-x unchanged, used the substitution 1-e-x =t2,

[tex]dx=\frac{2tdt}{1-t^2}[/tex]

I also got the logarithm of a fraction plus a constant times the square root. The problem can be solved, and you are right, it should rather be at "Calculus and Beyond"

ehild
 

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