Question

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)

Answer 1

Haha do I do trig substitution on this? I can't resist must go for it.

Answer 2

My calc could do this in two seconds

Answer 3

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)

Answer 4

and*

Answer 5

y'all wanna know the answer? the proof still remains...

Answer 6

Torturing students at U of M I see

Answer 7

Cinvestav, for masters and doctorates only

Answer 8

\[I=\int_0^\pi\frac{x\sin x}{1+\cos^2x}\text{d}x\]
\[x=\pi-t\]\[I=\int_\pi^0\frac{(\pi-t)\sin t}{1+\cos^2t}(-\text{d}t)=\int_0^\pi\frac{(\pi-t)\sin t}{1+\cos^2t}\text{d}t\]\[=\int_0^\pi\frac{\pi\sin t}{1+\cos^2t}\text{d}t-\int_0^\pi\frac{t\sin t}{1+\cos^2t}\text{d}t\]\[2I=\int_0^\pi\frac{\pi\sin t}{1+\cos^2t}\text{d}t=-\pi \ \tan^{-1} (\cos t) |_0^\pi=\frac{\pi^2}{4}\]\[I=\frac{\pi^2}{8}\]

Answer 9

emm...\[2I=\int_0^\pi\frac{\pi\sin t}{1+\cos^2t}\text{d}t=-\pi \ \tan^{-1} (\cos t) |_0^\pi=\frac{\pi^2}{2}\]\[I=\frac{\pi^2}{4}\]

Answer 10

@mukushla nailed it, as usual! Very similar to my approach, hats off :D

Answer 11

How do you people do that? :|

Answer 12

Stop being geniuses, or I'd kill myself one day for my ignorance. :P

Answer 13

It's called practice, as I always say...

Answer 14

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.

Answer 15

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