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anonymous
 one year ago
Evaluate the integral..
anonymous
 one year ago
Evaluate the integral..

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is arctan possible here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits _{\infty }^{\infty }\:\left(\frac{dx}{e^x+e^{x}}\right)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.5I would make it more obvious that is so by multiplying e^x on top and bottom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is in fact arctan in one step right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.5\[\int\limits_{}^{}\frac{ du}{u^2+1} \text{ where } u=e^x \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.5the lower limit of integration changes since as x goes to inf then u goes to 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles that is clever, i would have taking \[\lim_{t\to \infty}\tan^{1}(e^t)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about e^x do I have to bring that up?

freckles
 one year ago
Best ResponseYou've already chosen the best response.5who cares we aren't English snobs we are math snobs @satellite73

freckles
 one year ago
Best ResponseYou've already chosen the best response.5\[\int\limits_{\infty}^{\infty} \frac{1}{e^{x}+e^{x} } dx \\ \text{ multiply \top and bottom by } e^{x} \\ \int\limits_{\infty}^{\infty} \frac{e^{x}}{e^{2x}+1} dx \\ \text{ \let } u=e^{x} \\ du=e^{x} dx \\ \\ \int\limits_0^\infty \frac{du }{u^2+1}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.5@arvnoodle that have written out does it make more sense or still no sensE?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait I'll analyze it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why do you have to multiply e^x? Is that algebra or what, you do that if you want to bring something up to numerator?

freckles
 one year ago
Best ResponseYou've already chosen the best response.5\[\frac{e^x}{e^{x}}=1 \\ \text{ I multiplied by this fancy 1 to put it in the obvious } \arctan \text{ integral form }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I noticed the lower limit changed to 0 why is that

freckles
 one year ago
Best ResponseYou've already chosen the best response.5as x approaches inf e^x approaches 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay tried it in my calcu it reaches 0 so that make sense

freckles
 one year ago
Best ResponseYou've already chosen the best response.5\[u=e^x \\ x \rightarrow \infty \text{ then } u=e^{x} \rightarrow \infty \\ x \rightarrow \infty \text{ then } u=e^{x} \rightarrow 0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.5dw:1438745936722:dw notice as x gets infinitely negative large the y values are getting close to 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits _0^{\infty }\:\frac{du}{u^2+1}\] sooo

freckles
 one year ago
Best ResponseYou've already chosen the best response.5you can integrate that correct? you actually gave a formula for that earlier and other people had mentioned something about the pretty arctan( ) thingy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{a}\tan^{1}\left(\frac{u}{a}\right)+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is this the one? is my formula right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.5but a is 1 and we don't need the C because we have a definite integral

freckles
 one year ago
Best ResponseYou've already chosen the best response.5\[\int\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)_0^{z}]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay so arctan(e^x) ? with limits of 0 and inf

freckles
 one year ago
Best ResponseYou've already chosen the best response.5well if you change it back to arctan(e^(x)) then you need to go back to the original limits

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So will the final answer be infinity?

freckles
 one year ago
Best ResponseYou've already chosen the best response.5\[\int\limits\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)_0^{z}] \\ =\lim_{z \rightarrow \infty}[\arctan(z)\arctan(0)] \\ =\lim_{z \rightarrow \infty}\arctan(z)\] 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
 one year ago
Best ResponseYou've already chosen the best response.5so 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
 one year ago
Best ResponseYou've already chosen the best response.5anyways 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
 one year ago
Best ResponseYou've already chosen the best response.0okay let me get this cllear for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\tan^{1}\left(\frac{e^x}{1}\right)\right]\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.5if we put it back in terms of x then we have to go back to the original limits

freckles
 one year ago
Best ResponseYou've already chosen the best response.5but you will have to break that integral up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh my this problem kind of hard lol sorry for not getting it quickly

freckles
 one year ago
Best ResponseYou've already chosen the best response.5\[\int\limits_{\infty}^{\infty} \frac{1}{e^x+e^{x} } dx \\ =\int\limits_{\infty}^0 \frac{1}{e^{x}+e^{x} } dx+\int\limits_0^\infty \frac{1}{e^{x}+e^{x} } dx \\ \text{ check to see both integrals converge } \] 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
 one year ago
Best ResponseYou've already chosen the best response.5\[= \lim_{u \rightarrow \infty}[\arctan(e^{x})_u^0]+\lim_{z \rightarrow \infty}[\arctan(e^{x})_0^z]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm unsatisfied because by my logic the e^x reaches inf it means the answer is inf

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it doesn't make sense pi/2 to me but ohwell

freckles
 one year ago
Best ResponseYou've already chosen the best response.5as x gets large then e^x gets large then arctan(getting super large)=?

freckles
 one year ago
Best ResponseYou've already chosen the best response.5dw:1438747311900:dw look at the graph of y=arctan(x) as x gets supper large y approaches pi/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh I get by graph!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh my does it mean I need to graph when I'm answering this kind of question, kind of tedious lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but okay thankyou tho

freckles
 one year ago
Best ResponseYou've already chosen the best response.5well you eventually won't need to graph every thing you will just remember the end behavior of most graphs

freckles
 one year ago
Best ResponseYou've already chosen the best response.5the graph was more of me trying to convince you that it is pi/2 it is not needed in the answer

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0as 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
 one year ago
Best ResponseYou've already chosen the best response.0I got it now I got now!! by chain rule
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