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TuringTest
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Fun integral that seems to have gotten deleted, so I'm posting it again.\[\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx\]
 2 years ago
 2 years ago
TuringTest Group Title
Fun integral that seems to have gotten deleted, so I'm posting it again.\[\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx\]
 2 years ago
 2 years ago

This Question is Closed

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
@Nodata this is from the cinvestav exam, and it was one I could remember how to do. @mukushla perhaps you wanna have a try?
 2 years ago

jk_16 Group TitleBest ResponseYou've already chosen the best response.0
your killing me hahah
 2 years ago

mandonut Group TitleBest ResponseYou've already chosen the best response.0
i think i got it
 2 years ago

mandonut Group TitleBest ResponseYou've already chosen the best response.0
u integrate xsinx and sinx/cos^2x separetely
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
not how I did it, but there more than one way to skin a cat
 2 years ago

mandonut Group TitleBest ResponseYou've already chosen the best response.0
i read it upside down
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
I'm gonna leave this up for quite a while before giving any hints :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
wow I totally didn't mean to close this question I'll repost it :S
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
@TuringTest i might have it... although i might well have made a mistake. i have it coming out as a periodic integral after a substitution
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
ill post what i have
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
\[u = cosx \]\[\frac{du}{dx} = sinx\]\[dx = \frac{du}{sinx}\] let our integral = I \[I =  \int\limits^{1}_{1} \frac{\cos^{1}u}{1+u^2} du\]\[I = \int\limits^{1}_{1} \frac{\cos^{1}u}{1+u^2} du\]\[f'(u) = \frac{1}{1+u^2} , g(u) = \cos^{1}(u)\]\[f(u) = \tan^{1}u \text{ , }g'(u) = \frac{1}{\sqrt{1u^2}}\] \[\int\limits{f' \times g } \text{ } du = f \times g  \int\limits{f \times g'} \text{ } du\] \[I = [\tan^{1}(u) \times \cos^{1}(u)]^{1}_{1}  \int\limits{\frac{\tan^{1}u}{\sqrt{1u^2}}}du\]\[I = \frac{\pi^2}{4}  \int\limits{\frac{\tan^{1}u}{\sqrt{1u^2}}}du\] \[ \int\limits{\frac{\tan^{1}u}{\sqrt{1u^2}}}du = [\tan^{1}u \times \cos^{1}]^{1}_{1}  \int\limits{\frac{\cos^{1}u}{1+u^2}du}\] therefore \[I = \frac{\pi^2}{4}  I\]\[I= \frac{\pi^2}{8}\] if this is completely wrong my excuse is tiredness!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
oh you are sooooooo close! let me see if I can find your apparently minor mistake....
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
ah pellet i think i see a mistake already
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
last couple of lines i think i made a sign error?
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
aaah crap i think i made a sign error, and without it i have I = I
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
you have it right up until \[\int{\tan^{1}u\over\sqrt{1u^2}}du\]though you did it very differently than me there is a trick to this integral, would you like a hint?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
I personally think you overcomplicated things by leaving them in terms of u by the way, so it took me a while to decipher your work, but it is correct :)
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
so up until i tried to integrate by parts?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
the integration by parts was correct this setup for the integral is correct, though it looks much differently than mine as you left it in terms of u and I kept mine in x
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
you have done everything right, but you need to integrate this last guy....
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
ah ok, so i just evaluated that last guy wrong
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
ugh im tired, ill have one last crack at it :D
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
also, here's one in return: \[\int\limits^{1}_{0}{\frac{x}{x^3 + 1}}dx\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
thanks I'll write it down, I'm leaving soon
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
let me know if you want a hint on the other one...
 2 years ago

eigenschmeigen Group TitleBest ResponseYou've already chosen the best response.1
no hints, but i'll have to continue it tommorow :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
good spirit :) see ya!
 2 years ago
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