Evaluate the integral..

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Evaluate the integral..

Mathematics
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
is arctan possible here
\[\int\limits _{-\infty }^{\infty }\:\left(\frac{dx}{e^x+e^{-x}}\right)\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question