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

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TuringTestBest 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?
 one year ago

mandonutBest ResponseYou've already chosen the best response.0
u integrate xsinx and sinx/cos^2x separetely
 one year ago

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

mandonutBest ResponseYou've already chosen the best response.0
i read it upside down
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
I'm gonna leave this up for quite a while before giving any hints :)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
wow I totally didn't mean to close this question I'll repost it :S
 one year ago

eigenschmeigenBest 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
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
ill post what i have
 one year ago

eigenschmeigenBest 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!
 one year ago

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

eigenschmeigenBest ResponseYou've already chosen the best response.1
ah pellet i think i see a mistake already
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
last couple of lines i think i made a sign error?
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
aaah crap i think i made a sign error, and without it i have I = I
 one year ago

TuringTestBest 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?
 one year ago

TuringTestBest 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 :)
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
so up until i tried to integrate by parts?
 one year ago

TuringTestBest 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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
you have done everything right, but you need to integrate this last guy....
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
ah ok, so i just evaluated that last guy wrong
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
ugh im tired, ill have one last crack at it :D
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.2
thanks I'll write it down, I'm leaving soon
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
let me know if you want a hint on the other one...
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
no hints, but i'll have to continue it tommorow :)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
good spirit :) see ya!
 one year ago
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