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TuringTest
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Fun integral that seems to have gotten deleted, so I'm posting it again.\[\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx\](yes I am an idiot, I accidentally closed the other post)
 one year ago
 one year ago
TuringTest Group Title
Fun integral that seems to have gotten deleted, so I'm posting it again.\[\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx\](yes I am an idiot, I accidentally closed the other post)
 one year ago
 one year ago

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Kainui Group TitleBest ResponseYou've already chosen the best response.0
Haha do I do trig substitution on this? I can't resist must go for it.
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
My calc could do this in two seconds
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
actually the answer on the test is given, ad I think wolf can do it to the problem on the test is to prove the result is what it is (it's from the entrance exam to a university here in Mexico)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
y'all wanna know the answer? the proof still remains...
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Torturing students at U of M I see
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
Cinvestav, for masters and doctorates only
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.5
\[I=\int_0^\pi\frac{x\sin x}{1+\cos^2x}\text{d}x\] \[x=\pit\]\[I=\int_\pi^0\frac{(\pit)\sin t}{1+\cos^2t}(\text{d}t)=\int_0^\pi\frac{(\pit)\sin t}{1+\cos^2t}\text{d}t\]\[=\int_0^\pi\frac{\pi\sin t}{1+\cos^2t}\text{d}t\int_0^\pi\frac{t\sin t}{1+\cos^2t}\text{d}t\]\[2I=\int_0^\pi\frac{\pi\sin t}{1+\cos^2t}\text{d}t=\pi \ \tan^{1} (\cos t) _0^\pi=\frac{\pi^2}{4}\]\[I=\frac{\pi^2}{8}\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.5
emm...\[2I=\int_0^\pi\frac{\pi\sin t}{1+\cos^2t}\text{d}t=\pi \ \tan^{1} (\cos t) _0^\pi=\frac{\pi^2}{2}\]\[I=\frac{\pi^2}{4}\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
@mukushla nailed it, as usual! Very similar to my approach, hats off :D
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
How do you people do that? :
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Stop being geniuses, or I'd kill myself one day for my ignorance. :P
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
It's called practice, as I always say...
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.0
This just seems like guessing what the solutions to a quadratic equation are or what the integrating factor of an exact differential equation is without actually working it out. How do you practice this? I don't remember this in Cal 1 or 2, but I'm damn interested. Maybe I'm not conceptualizing integrals in their simplest possible form... if that makes sense.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
The best reply to that is 3 hints... hint one: integration by parts let me know how that goes and I will give more hints if necessary
 one year ago
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