Square Root Indefinite Integral

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  • #1
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Hello everyone..

Homework Statement


∫√((1+(e^-x))^2)dx




2. The attempt at a solution
I first tried to do a u sub and then attempt a trig sub however I cant do anything with the e^-x left in the u sub. Does anyone have another way I can integrate this thing??

Thank you for any suggestions/help!
 
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Answers and Replies

  • #2
Ray Vickson
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Hello everyone..

Homework Statement


∫√((1+e^-x)^2)dx




2. The attempt at a solution
I first tried to do a u sub and then attempt a trig sub however I cant do anything with the e^-x left in the u sub. Does anyone have another way I can integrate this thing??

Thank you for any suggestions/help!
Before doing anything, try computing the square root.
 
  • #3
B18
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The integral is within an arc length problem if it matters. However when i integrate (1+(e^-x)^2)^1/2 from 0 to 2 i get roughly 2.2214187 and when i integrate 1+e^-x from 0 to 2 i get 2.8647.
Am I allowed to square root 1+(e^-x)^2 and receive 1+e^-x?
 
  • #4
B18
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I'm sorry I forgot the parenthesis around the (e^-x)^2 it should be square root of ((1+((e^-x)^2)))
 
  • #5
SammyS
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The integral is within an arc length problem if it matters. However when i integrate (1+(e^-x)^2)^1/2 from 0 to 2 i get roughly 2.2214187 and when i integrate 1+e^-x from 0 to 2 i get 2.8647.
Am I allowed to square root 1+(e^-x)^2 and receive 1+e^-x?
I'm sorry I forgot the parenthesis around the (e^-x)^2 it should be square root of ((1+((e^-x)^2)))
Well that changes everything.


Then, of course, [itex]\displaystyle \ \sqrt{1+\left(e^{-x}\right)^2}\ne 1+e^{-x}\ .\ [/itex] You wouldn't expect the two integrals to give the same result.
 
  • #6
Ray Vickson
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I'm sorry I forgot the parenthesis around the (e^-x)^2 it should be square root of ((1+((e^-x)^2)))
Using too many brackets is almost as confusing as not using enough. Do you mean (a): ##\sqrt{1+(e^{-x})^2}## or (b): ##\sqrt{(1 + e^{-x})^2}##? If you mean (a) you can write it as sqrt(1 + (e^{-x})^2); if you mean (b) you can write it as sqrt((1 + e^{-x})^2). You can even write e^-x without the {} bracket, as you have done, but some people would argue with that.

I thought in your OP you meant (b), and that is why I suggested you do the square root first. In (a) you can still simplify it usefully before tackling the integral.
 
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  • #7
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Well that changes everything.


Then, of course, [itex]\displaystyle \ \sqrt{1+\left(e^{-x}\right)^2}\ne 1+e^{-x}\ .\ [/itex] You wouldn't expect the two integrals to give the same result.
your right, so is there a specific technique of integration I should use for this integral?
 
  • #8
B18
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Using too many brackets is almost as confusing as not using enough. Do you mean (a): ##\sqrt{1+(e^{-x})^2}## or (b): ##\sqrt{(1 + e^{-x})^2}##? If you mean (a) you can write it as sqrt(1 + (e^{-x})^2); if you mean (b) you can write it as sqrt((1 + e^{-x})^2). You can even write e^-x without the {} bracket, as you have done, but some people would argue with that.

I thought in your OP you meant (a), and that is why I suggested you do the square root first. In (b) you can still simplify it usefully before tackling the integral.
A is what the integral i must find looks like. I truly wish it had looked like B though...
 
  • #9
B18
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Anyone have an idea?
 
  • #10
SammyS
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Anyone have an idea?
Looking at the results from WolframAlpha, the indefinite integral is messy, but not impossible.
 
  • #11
B18
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I'm still attempting to solve it but haven't gotten anywhere yet.
 
  • #12
B18
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If I substituted u= e^x in any integral and the e^x didn't cross out can I just let it equal u and multiply 1/u into the original integral?? Or is that illegal?
 
  • #13
B18
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After three substitutions and a partial fraction I have the answer and it matches wolfram alpha. Thanks for the help
 
  • #14
So, just to clarify, the integral is

##\int \sqrt{1+e^{-2x}}\;dx##?
 
  • #15
B18
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Yes that's correct
 
  • #16
Alright, well here's what I did:

##\displaystyle\int\sqrt{1+e^{-2x}}\;dx\\
\displaystyle\int\sqrt{1+e^{-2x}}\cdot\dfrac{\sqrt{1+e^{-2x}}}{\sqrt{1+e^{-2x}}}\;dx\\
\displaystyle\int \dfrac{1+e^{-2x}}{\sqrt{1+e^{-2x}}}\;dx\\
\displaystyle\int \dfrac{1+e^{-2x}}{\sqrt{1+e^{-2x}}}\cdot\dfrac{\sqrt{1-e^{-2x}}}{\sqrt{1-e^{-2x}}}\;dx\\
\displaystyle\int \dfrac{\left(1+e^{-2x}\right)\sqrt{1-e^{-2x}}}{1-e^{-2x}}\;dx##

As a substitution, let ##u=e^{-x},\;du=-e^{-x}dx\Rightarrow dx=-\dfrac{du}{u}##.

##\displaystyle-\int \dfrac{\left(1+u^2\right)\sqrt{1-u^2}}{1-u^2}\;\dfrac{du}{u}##

Then, a trig substitution with ##u = \sin t,\;du=\cos t\;dt##.

##\displaystyle-\int \dfrac{\left(1+\sin^2t\right)\sqrt{1-\sin^2t}}{\sin t\left(1-\sin^2t\right)}\cos t\;dt##

Something tells me there may have been a slightly shorter route in the first set of rewriting, though...
 
  • #17
Actually, I take it back. After looking over the first few lines, it doesn't look like stopping after one of the earlier steps gives you an easy integral to work with.
 
  • #18
B18
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Another way you could possibly do this would be a u= e^-x then a trig sub (tan)
 
  • #19
Ray Vickson
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Another way you could possibly do this would be a u= e^-x then a trig sub (tan)
For ##\int \sqrt{1+e^{-2x}}\, dx,## substitute ##e^{-2x} = u.##
 
  • #20
B18
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I would leave the e^-x squared that way the trig sub would work after u=e^-x
 
  • #21
Math_QED
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Just another approach. Consider the substitution x = ln u. So dx = 1/u du

The integral becomes: ##\int \sqrt{1+u^{-2}}/u du##. From here, the integral should be pretty straight-forward with some trig substitutions.
 
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  • #22
SammyS
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I would leave the e^-x squared that way the trig sub would work after u=e^-x
You have just responded to a thread that's over 3 years old !
 

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