Evaluate the integral..

- anonymous

Evaluate the integral..

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- anonymous

##### 1 Attachment

- anonymous

is arctan possible here

- anonymous

\[\int\limits _{-\infty }^{\infty }\:\left(\frac{dx}{e^x+e^{-x}}\right)\]

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## More answers

- Zale101

Yes, it is possible

- freckles

I would make it more obvious that is so by multiplying e^x on top and bottom

- anonymous

it is in fact arctan in one step right?

- anonymous

The problem is I don't know how first time seeing such problem\[\int\limits \frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C \] took this from my formula ntbk

- freckles

\[\int\limits_{}^{}\frac{ du}{u^2+1} \text{ where } u=e^x \]

- freckles

the lower limit of integration changes since as x goes to -inf
then u goes to 0

- anonymous

@freckles that is clever, i would have taking
\[\lim_{t\to -\infty}\tan^{-1}(e^t)\]

- anonymous

*taken

- anonymous

what about e^-x do I have to bring that up?

- freckles

who cares we aren't English snobs we are math snobs @satellite73

- anonymous

\[\frac{e^x}{e^x}\]

- freckles

\[\int\limits_{-\infty}^{\infty} \frac{1}{e^{x}+e^{-x} } dx \\ \text{ multiply \top and bottom by } e^{x} \\ \int\limits_{-\infty}^{\infty} \frac{e^{x}}{e^{2x}+1} dx \\ \text{ \let } u=e^{x} \\ du=e^{x} dx \\ \\ \int\limits_0^\infty \frac{du }{u^2+1}\]

- freckles

@arvnoodle that have written out does it make more sense or still no sensE?

- anonymous

Wait I'll analyze it

- anonymous

Why do you have to multiply e^x? Is that algebra or what, you do that if you want to bring something up to numerator?

- Zale101

|dw:1438745594477:dw|

- freckles

\[\frac{e^x}{e^{x}}=1 \\ \text{ I multiplied by this fancy 1 to put it in the obvious } \arctan \text{ integral form }\]

- freckles

yes that is algebra

- anonymous

I noticed the lower limit changed to 0 why is that

- freckles

as x approaches -inf
e^x approaches 0

- anonymous

Oh okay tried it in my calcu it reaches 0 so that make sense

- freckles

\[u=e^x \\ x \rightarrow \infty \text{ then } u=e^{x} \rightarrow \infty \\ x \rightarrow -\infty \text{ then } u=e^{x} \rightarrow 0\]

- freckles

|dw:1438745936722:dw|
notice as x gets infinitely negative large the y values are getting close to 0

- anonymous

\[\int\limits _0^{\infty }\:\frac{du}{u^2+1}\] sooo

- freckles

you can integrate that correct?
you actually gave a formula for that earlier
and other people had mentioned something about the pretty arctan( ) thingy

- anonymous

\[\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C\]

- anonymous

is this the one? is my formula right?

- freckles

but a is 1
and we don't need the C because we have a definite integral

- freckles

\[\int\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)|_0^{z}]\]

- anonymous

Oh okay so arctan(e^x) ? with limits of 0 and inf

- freckles

well if you change it back to arctan(e^(x))
then you need to go back to the original limits

- anonymous

So will the final answer be infinity?

- freckles

no

- freckles

\[\int\limits\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)|_0^{z}] \\ =\lim_{z \rightarrow \infty}[\arctan(z)-\arctan(0)] \\ =\lim_{z \rightarrow \infty}\arctan(z)\]
y=tan(x) on (-pi/2,pi/2) looks like:
|dw:1438746457408:dw|
so y=arctan(x) on (-inf,inf) looks like:
|dw:1438746492037:dw|

- freckles

so looking at this y=arctan(x) on (-inf,inf) in your graphing calculator you should see as x gets super large then the y values are approaching y=pi/2

- freckles

anyways can you tell me the following limit:
\[\lim_{z \rightarrow \infty}\arctan(z)\]
it should be obvious from the graph and what I just said that this is ....

- anonymous

okay let me get this cllear for me

- anonymous

\[\left[\tan^{-1}\left(\frac{e^x}{1}\right)\right]\]

- anonymous

that's right? ^

- freckles

if we put it back in terms of x then we have to go back to the original limits

- anonymous

okay -inf to +inf

- freckles

which is fine

- freckles

but you will have to break that integral up

- anonymous

the -inf turns to 0

- anonymous

Hmmm break?

- anonymous

oh my this problem kind of hard lol sorry for not getting it quickly

- freckles

\[\int\limits_{-\infty}^{\infty} \frac{1}{e^x+e^{-x} } dx \\ =\int\limits_{-\infty}^0 \frac{1}{e^{x}+e^{-x} } dx+\int\limits_0^\infty \frac{1}{e^{x}+e^{-x} } dx \\ \text{ check to see both integrals converge } \]
which we know they will because we already seen the answer another way
so the first integral will converge to the sum of the other two integrals in this case

- freckles

\[= \lim_{u \rightarrow -\infty}[\arctan(e^{x})|_u^0]+\lim_{z \rightarrow \infty}[\arctan(e^{x})|_0^z]\]

- anonymous

the answer is pi/2

- freckles

yes

- anonymous

I'm unsatisfied because by my logic the e^x reaches inf it means the answer is inf

- anonymous

it doesn't make sense pi/2 to me but ohwell

- freckles

as x gets large then e^x gets large
then arctan(getting super large)=?

- freckles

|dw:1438747311900:dw|
look at the graph of y=arctan(x)
as x gets supper large y approaches pi/2

- freckles

super!

- anonymous

oh I get by graph!!!

- anonymous

oh my does it mean I need to graph when I'm answering this kind of question, kind of tedious lol

- anonymous

but okay thankyou tho

- freckles

well you eventually won't need to graph every thing
you will just remember the end behavior of most graphs

- freckles

the graph was more of me trying to convince you that it is pi/2
it is not needed in the answer

- UnkleRhaukus

as the opposite side of a right angled triangle approaches infinitely greater in length than the adjacent side, the considered angle approaches 90° = π/2

- anonymous

I got it now I got now!! by chain rule

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