## TuringTest 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

1. Kainui

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

2. Hero

My calc could do this in two seconds

3. TuringTest

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)

4. TuringTest

and*

5. TuringTest

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

6. Hero

Torturing students at U of M I see

7. TuringTest

Cinvestav, for masters and doctorates only

8. mukushla

$I=\int_0^\pi\frac{x\sin x}{1+\cos^2x}\text{d}x$ $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}$

9. mukushla

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}$

10. TuringTest

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

11. ParthKohli

How do you people do that? :|

12. ParthKohli

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

13. TuringTest

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

14. Kainui

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.

15. TuringTest

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