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

- TuringTest

- schrodinger

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- TuringTest

@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?

- anonymous

your killing me hahah

- anonymous

i think i got it

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## More answers

- anonymous

u integrate xsinx and sinx/cos^2x separetely

- TuringTest

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

- anonymous

lol nvm

- anonymous

i read it upside down

- TuringTest

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

- TuringTest

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

- anonymous

@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

- anonymous

ill post what i have

- anonymous

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

- TuringTest

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

- anonymous

ah pellet i think i see a mistake already

- anonymous

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

- anonymous

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

- TuringTest

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?

- TuringTest

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 :)

- anonymous

so up until i tried to integrate by parts?

- TuringTest

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

- TuringTest

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

- anonymous

ah ok, so i just evaluated that last guy wrong

- TuringTest

yes

- anonymous

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

- anonymous

also, here's one in return:
\[\int\limits^{1}_{0}{\frac{x}{x^3 + 1}}dx\]

- TuringTest

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

- TuringTest

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

- anonymous

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

- TuringTest

good spirit :)
see ya!

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