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

Mathematics
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@No-data this is from the cinvestav exam, and it was one I could remember how to do. @mukushla perhaps you wanna have a try?
your killing me hahah
i think i got it

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Other answers:

u integrate xsinx and sinx/cos^2x separetely
not how I did it, but there more than one way to skin a cat
lol nvm
i read it upside down
I'm gonna leave this up for quite a while before giving any hints :)
wow I totally didn't mean to close this question I'll repost it :S
@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
ill post what i have
\[u = cosx \]\[\frac{du}{dx} = -sinx\]\[dx = \frac{du}{-sinx}\] let our integral = I \[I = - \int\limits^{-1}_{1} \frac{\cos^{-1}u}{1+u^2} du\]\[I = \int\limits^{1}_{-1} \frac{\cos^{-1}u}{1+u^2} du\]\[f'(u) = \frac{1}{1+u^2} , g(u) = \cos^{-1}(u)\]\[f(u) = \tan^{-1}u \text{ , }g'(u) = \frac{1}{\sqrt{1-u^2}}\] \[\int\limits{f' \times g } \text{ } du = f \times g - \int\limits{f \times g'} \text{ } du\] \[I = [\tan^{-1}(u) \times \cos^{-1}(u)]^{1}_{-1} - \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du\]\[I = \frac{\pi^2}{4} - \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du\] \[ \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du = [\tan^{-1}u \times \cos^{-1}]^{1}_{-1} - \int\limits{\frac{\cos^{-1}u}{1+u^2}du}\] therefore \[I = \frac{\pi^2}{4} - I\]\[I= \frac{\pi^2}{8}\] if this is completely wrong my excuse is tiredness!
oh you are sooooooo close! let me see if I can find your apparently minor mistake....
ah pellet i think i see a mistake already
last couple of lines i think i made a sign error?
aaah crap i think i made a sign error, and without it i have I = I
you have it right up until \[\int{\tan^{-1}u\over\sqrt{1-u^2}}du\]though you did it very differently than me there is a trick to this integral, would you like a hint?
I personally think you over-complicated 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 :)
so up until i tried to integrate by parts?
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
you have done everything right, but you need to integrate this last guy....
ah ok, so i just evaluated that last guy wrong
yes
ugh im tired, ill have one last crack at it :D
also, here's one in return: \[\int\limits^{1}_{0}{\frac{x}{x^3 + 1}}dx\]
thanks I'll write it down, I'm leaving soon
let me know if you want a hint on the other one...
no hints, but i'll have to continue it tommorow :)
good spirit :) see ya!

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