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

This Question is Closed

TuringTest
 3 years ago
Best 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your killing me hahah

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u integrate xsinx and sinx/cos^2x separetely

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i read it upside down

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[u = cosx \]\[\frac{du}{dx} = sinx\]\[dx = \frac{du}{sinx}\] 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!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ah pellet i think i see a mistake already

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0last couple of lines i think i made a sign error?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0aaah crap i think i made a sign error, and without it i have I = I

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2you 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?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2I 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 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so up until i tried to integrate by parts?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2the 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ah ok, so i just evaluated that last guy wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ugh im tired, ill have one last crack at it :D

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no hints, but i'll have to continue it tommorow :)

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