Question

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

Answer 1

@No-data this is from the cinvestav exam, and it was one I could remember how to do.
@mukushla perhaps you wanna have a try?

Answer 2

your killing me hahah

Answer 3

i think i got it

Answer 4

u integrate xsinx and sinx/cos^2x separetely

Answer 5

not how I did it, but there more than one way to skin a cat

Answer 6

lol nvm

Answer 7

i read it upside down

Answer 8

I'm gonna leave this up for quite a while before giving any hints :)

Answer 9

wow I totally didn't mean to close this question
I'll repost it :S

Answer 10

@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

Answer 11

ill post what i have

Answer 12

\[u = cosx \]\[\frac{du}{dx} = -sinx\]\[dx = \frac{du}{-sinx}\]

let our integral = I

\[I = - \int\limits^{-1}_{1} \frac{\cos^{-1}u}{1+u^2} du\]\[I = \int\limits^{1}_{-1} \frac{\cos^{-1}u}{1+u^2} du\]\[f'(u) = \frac{1}{1+u^2} , g(u) = \cos^{-1}(u)\]\[f(u) = \tan^{-1}u \text{ , }g'(u) = \frac{1}{\sqrt{1-u^2}}\]
\[\int\limits{f' \times g } \text{ } du = f \times g - \int\limits{f \times g'} \text{ } du\]


\[I = [\tan^{-1}(u) \times \cos^{-1}(u)]^{1}_{-1} - \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du\]\[I = \frac{\pi^2}{4} - \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du\]

\[ \int\limits{\frac{\tan^{-1}u}{\sqrt{1-u^2}}}du = [\tan^{-1}u \times \cos^{-1}]^{1}_{-1} - \int\limits{\frac{\cos^{-1}u}{1+u^2}du}\]

therefore
\[I = \frac{\pi^2}{4} - I\]\[I= \frac{\pi^2}{8}\]

if this is completely wrong my excuse is tiredness!

Answer 13

oh you are sooooooo close!
let me see if I can find your apparently minor mistake....

Answer 14

ah pellet i think i see a mistake already

Answer 15

last couple of lines i think i made a sign error?

Answer 16

aaah crap i think i made a sign error, and without it i have I = I

Answer 17

you have it right up until \[\int{\tan^{-1}u\over\sqrt{1-u^2}}du\]though you did it very differently than me
there is a trick to this integral, would you like a hint?

Answer 18

I personally think you over-complicated 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 :)

Answer 19

so up until i tried to integrate by parts?

Answer 20

the 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

Answer 21

you have done everything right, but you need to integrate this last guy....

Answer 22

ah ok, so i just evaluated that last guy wrong

Answer 23

yes

Answer 24

ugh im tired, ill have one last crack at it :D

Answer 25

also, here's one in return:

\[\int\limits^{1}_{0}{\frac{x}{x^3 + 1}}dx\]

Answer 26

thanks I'll write it down, I'm leaving soon

Answer 27

let me know if you want a hint on the other one...

Answer 28

no hints, but i'll have to continue it tommorow :)

Answer 29

good spirit :)
see ya!