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

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Kainui
 2 years ago
Best ResponseYou've already chosen the best response.0Haha do I do trig substitution on this? I can't resist must go for it.

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0My calc could do this in two seconds

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0actually 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)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0y'all wanna know the answer? the proof still remains...

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0Torturing students at U of M I see

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0Cinvestav, for masters and doctorates only

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

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

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0How do you people do that? :

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0It's called practice, as I always say...

Kainui
 2 years ago
Best ResponseYou've already chosen the best response.0This 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.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0The 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
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