Square Root Indefinite Integral

[B18]

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 can't 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!

 

[Ray Vickson]

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 can't 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.

 

[B18]

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?

 

[B18]

I'm sorry I forgot the parenthesis around the (e^-x)^2 it should be square root of ((1+((e^-x)^2)))

 

[SammyS]

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, $\displaystyle \ \sqrt{1+\left(e^{-x}\right)^2}\ne 1+e^{-x}\ .\$ You wouldn't expect the two integrals to give the same result.

 

[Ray Vickson]

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.

 

[B18]

Well that changes everything.


Then, of course, $\displaystyle \ \sqrt{1+\left(e^{-x}\right)^2}\ne 1+e^{-x}\ .\$ 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?

 

[B18]

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...

 

[B18]

Anyone have an idea?

 

[SammyS]

Anyone have an idea?

Looking at the results from WolframAlpha, the indefinite integral is messy, but not impossible.

 

[B18]

I'm still attempting to solve it but haven't gotten anywhere yet.

 

[B18]

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?

 

[B18]

After three substitutions and a partial fraction I have the answer and it matches wolfram alpha. Thanks for the help

 

[SithsNGiggles]

So, just to clarify, the integral is

$\int \sqrt{1+e^{-2x}}\;dx$?

 

[B18]

Yes that's correct

 

[SithsNGiggles]

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...

 

[SithsNGiggles]

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.

 

[B18]

Another way you could possibly do this would be a u= e^-x then a trig sub (tan)

 

[Ray Vickson]

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 $e^{-2x} = u.$

 

[B18]

I would leave the e^-x squared that way the trig sub would work after u=e^-x

 

[member 587159]

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.

 

[SammyS]

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 !