anonymous
  • anonymous
Evaluate the integral..
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
is arctan possible here
anonymous
  • anonymous
\[\int\limits _{-\infty }^{\infty }\:\left(\frac{dx}{e^x+e^{-x}}\right)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Zale101
  • Zale101
Yes, it is possible
freckles
  • freckles
I would make it more obvious that is so by multiplying e^x on top and bottom
anonymous
  • anonymous
it is in fact arctan in one step right?
anonymous
  • 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
  • freckles
\[\int\limits_{}^{}\frac{ du}{u^2+1} \text{ where } u=e^x \]
freckles
  • freckles
the lower limit of integration changes since as x goes to -inf then u goes to 0
anonymous
  • anonymous
@freckles that is clever, i would have taking \[\lim_{t\to -\infty}\tan^{-1}(e^t)\]
anonymous
  • anonymous
*taken
anonymous
  • anonymous
what about e^-x do I have to bring that up?
freckles
  • freckles
who cares we aren't English snobs we are math snobs @satellite73
anonymous
  • anonymous
\[\frac{e^x}{e^x}\]
freckles
  • 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
  • freckles
@arvnoodle that have written out does it make more sense or still no sensE?
anonymous
  • anonymous
Wait I'll analyze it
anonymous
  • 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
  • Zale101
|dw:1438745594477:dw|
freckles
  • 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
  • freckles
yes that is algebra
anonymous
  • anonymous
I noticed the lower limit changed to 0 why is that
freckles
  • freckles
as x approaches -inf e^x approaches 0
anonymous
  • anonymous
Oh okay tried it in my calcu it reaches 0 so that make sense
freckles
  • 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
  • freckles
|dw:1438745936722:dw| notice as x gets infinitely negative large the y values are getting close to 0
anonymous
  • anonymous
\[\int\limits _0^{\infty }\:\frac{du}{u^2+1}\] sooo
freckles
  • 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
  • anonymous
\[\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C\]
anonymous
  • anonymous
is this the one? is my formula right?
freckles
  • freckles
but a is 1 and we don't need the C because we have a definite integral
freckles
  • freckles
\[\int\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)|_0^{z}]\]
anonymous
  • anonymous
Oh okay so arctan(e^x) ? with limits of 0 and inf
freckles
  • freckles
well if you change it back to arctan(e^(x)) then you need to go back to the original limits
anonymous
  • anonymous
So will the final answer be infinity?
freckles
  • freckles
no
freckles
  • 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
  • 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
  • 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
  • anonymous
okay let me get this cllear for me
anonymous
  • anonymous
\[\left[\tan^{-1}\left(\frac{e^x}{1}\right)\right]\]
anonymous
  • anonymous
that's right? ^
freckles
  • freckles
if we put it back in terms of x then we have to go back to the original limits
anonymous
  • anonymous
okay -inf to +inf
freckles
  • freckles
which is fine
freckles
  • freckles
but you will have to break that integral up
anonymous
  • anonymous
the -inf turns to 0
anonymous
  • anonymous
Hmmm break?
anonymous
  • anonymous
oh my this problem kind of hard lol sorry for not getting it quickly
freckles
  • 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
  • freckles
\[= \lim_{u \rightarrow -\infty}[\arctan(e^{x})|_u^0]+\lim_{z \rightarrow \infty}[\arctan(e^{x})|_0^z]\]
anonymous
  • anonymous
the answer is pi/2
freckles
  • freckles
yes
anonymous
  • anonymous
I'm unsatisfied because by my logic the e^x reaches inf it means the answer is inf
anonymous
  • anonymous
it doesn't make sense pi/2 to me but ohwell
freckles
  • freckles
as x gets large then e^x gets large then arctan(getting super large)=?
freckles
  • freckles
|dw:1438747311900:dw| look at the graph of y=arctan(x) as x gets supper large y approaches pi/2
freckles
  • freckles
super!
anonymous
  • anonymous
oh I get by graph!!!
anonymous
  • anonymous
oh my does it mean I need to graph when I'm answering this kind of question, kind of tedious lol
anonymous
  • anonymous
but okay thankyou tho
freckles
  • freckles
well you eventually won't need to graph every thing you will just remember the end behavior of most graphs
freckles
  • freckles
the graph was more of me trying to convince you that it is pi/2 it is not needed in the answer
UnkleRhaukus
  • 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
  • anonymous
I got it now I got now!! by chain rule

Looking for something else?

Not the answer you are looking for? Search for more explanations.