determine if the improper integral of 4/sin(x)(sqrt(1-x^2) from 0 to 1converges and if so at what value?

- anonymous

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- Loser66

\[\int_0^1\frac{4}{sinx \sqrt{1-x^2}}~~~or~~~\int_0^1 \frac{4}{sinx}\sqrt{1-x^2}dx\] which one?

- anonymous

the first one

- Loser66

I am sorry, I really don't know,

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## More answers

- anonymous

thats ok neither do i

- anonymous

this is very complicated stuff

- Loser66

@ash2326

- Psymon

\[\frac{ 4 }{ \sin(x)\sqrt{1-x ^{2}} }\]?
What loser tried to post shows as an error for me, so just want to confirm if this is it

- anonymous

yes also you take the integral of that from 0 to 1

- inkyvoyd

maybe integration by parts and trig sub?

- anonymous

how so?

- inkyvoyd

well it looks like your u and v' will be 1/sin x and 1/sqrt(1-x^2)

- inkyvoyd

you knnow how to find the integral of both expressions, but you really want v to be what I chose. try it out and lmk how it goes; I have to do homework myself unfortunately

- Psymon

Yeah, that does work out actually. I guess I was just double checking myself.

- inkyvoyd

btw that's my excuse for not doing the problem since I don't feel like it lol

- anonymous

ok i'll try that thanks

- Psymon

Mhm. You'll have to do integration by parts twice. It comes out pretty clean, though.

- inkyvoyd

protip if you're allowed to use a table for the trig sub, it's a pain in the arse

- inkyvoyd

@Psymon want a even simpler integration problem that's really hard to evaluate?

- Psymon

Lol, why would I want to do more unnecessary work? xD This one just turned out to be okay.

- inkyvoyd

does that integral even converge btw?>

- Psymon

Nope.

- inkyvoyd

LOL

- Psymon

But just saying that isnt enough, he still has to be able to work his way through it to determine that, lol.

- inkyvoyd

yeah guy42 use limits

- anonymous

it doesn't converge? are you sure?

- Psymon

Positive.

- anonymous

ok thanks

- anonymous

nope it does converge

- inkyvoyd

gg

- Psymon

Really? O.o

- anonymous

the not converge answer choice was wrong

- inkyvoyd

do you get another try?

- anonymous

3 more tries

- Psymon

I dont see how, haha.
\[\int\limits_{0}^{1}\frac{ 4 }{ \sin(x)\sqrt{1-x ^{2}} } \]
If thats the integral then Im trying to see how that actually converges x_x

- inkyvoyd

psymon go to wolfram alpha, and someone gimme a medal or inky be sad

- Psymon

wolfram goes all dead trying to do it

- inkyvoyd

kay
lemme try mathematica

- Psymon

Okay, got wolfram to do it
http://www.wolframalpha.com/input/?i=integrate+4%2F%28sinx*sqrt%281%2Bx^2%29%29+0+to+1&lk=4&num=1

- Psymon

Oh wow.....I think I got an answer x_x This is one funky integral, questionable accuracy like crazy.

- anonymous

i got the answer (pi^2)/2

- Psymon

And thats still not the answer I got, haha. Like seriously, wtf?

- inkyvoyd

ok, did you guys try out what I suggested with the by parts and all?

- Psymon

Yeah, I did.

- inkyvoyd

does it work?

- Psymon

I just gotta follow through with it. I guess I need to see if I can trust my pencil and paper more than wolfram

- anonymous

ok guys im gonna go thanks for your help

- Psymon

e

- inkyvoyd

if you take v'=csc x you get
Log[Sin[x/2]] Log[Tan[x/4]] + Log[1 - I Tan[x/4]] Log[Tan[x/4]] +
Log[1 + I Tan[x/4]] Log[Tan[x/4]] - 1/2 Log[Tan[x/4]]^2 +
Log[(-(1/2) - I/2) (-1 + Tan[x/4])] Log[-I + Tan[x/4]] +
Log[(-(1/2) + I/2) (-1 + Tan[x/4])] Log[I + Tan[x/4]] +
Log[-I + Tan[x/4]] Log[(1/2 - I/2) (1 + Tan[x/4])] +
Log[I + Tan[x/4]] Log[(1/2 + I/2) (1 + Tan[x/4])] -
Log[Tan[x/4]] Log[1 + Tan[x/4]] -
1/2 Log[Sin[x/2]] Log[1 - Tan[x/4]^2] +
1/2 Log[Tan[x/4]] Log[1 - Tan[x/4]^2] -
1/2 Log[-I + Tan[x/4]] Log[1 - Tan[x/4]^2] -
1/2 Log[I + Tan[x/4]] Log[1 - Tan[x/4]^2] -
1/2 Log[Sin[x/2]] Log[-1 + Tan[x/4]^2] +
1/2 Log[Tan[x/4]] Log[-1 + Tan[x/4]^2] -
1/2 Log[-I + Tan[x/4]] Log[-1 + Tan[x/4]^2] -
1/2 Log[I + Tan[x/4]] Log[-1 + Tan[x/4]^2] +
PolyLog[2, 1 - Tan[x/4]] +
PolyLog[2, 1/2 ((1 + I) - (1 - I) Tan[x/4])] -
PolyLog[2, -Tan[x/4]] + PolyLog[2, -I Tan[x/4]] +
PolyLog[2, I Tan[x/4]] + PolyLog[2, (-(1/2) - I/2) (I + Tan[x/4])] +
PolyLog[2, 1/2 ((1 + I) + (1 - I) Tan[x/4])] +
PolyLog[2,
1/2 ((1 - I) + (1 + I) Tan[x/4])] - (Log[
Cos[x/2]] (Log[2] Log[1 - Tan[x/4]] - 1/2 Log[1 - Tan[x/4]]^2 +
Log[Cos[x/4]^2] Log[Tan[x/4]] +
Log[1 - I Tan[x/4]] Log[Tan[x/4]] +
Log[1 + I Tan[x/4]] Log[Tan[x/4]] +
Log[(-(1/2) - I/2) (-1 + Tan[x/4])] Log[-I + Tan[x/4]] +
Log[(-(1/2) + I/2) (-1 + Tan[x/4])] Log[I + Tan[x/4]] +
Log[-I + Tan[x/4]] Log[(1/2 - I/2) (1 + Tan[x/4])] +
Log[I + Tan[x/4]] Log[(1/2 + I/2) (1 + Tan[x/4])] +
Log[2] Log[1 + Tan[x/4]] -
2 Log[1 - Tan[x/4]] Log[1 + Tan[x/4]] -
1/2 Log[1 + Tan[x/4]]^2 -
1/2 Log[Cos[x/4]^2] Log[1 - Tan[x/4]^2] -
1/2 Log[-I + Tan[x/4]] Log[1 - Tan[x/4]^2] -
1/2 Log[I + Tan[x/4]] Log[1 - Tan[x/4]^2] -
1/2 Log[Cos[x/4]^2] Log[-1 + Tan[x/4]^2] -
1/2 Log[-I + Tan[x/4]] Log[-1 + Tan[x/4]^2] -
1/2 Log[I + Tan[x/4]] Log[-1 + Tan[x/4]^2] -
PolyLog[2, 1/2 (1 - Tan[x/4])] +
PolyLog[2, 1/2 ((1 + I) - (1 - I) Tan[x/4])] -
PolyLog[2, -Tan[x/4]] + PolyLog[2, -I Tan[x/4]] +
PolyLog[2, I Tan[x/4]] - PolyLog[2, Tan[x/4]] +
PolyLog[2, (-(1/2) - I/2) (I + Tan[x/4])] -
PolyLog[2, 1/2 (1 + Tan[x/4])] +
PolyLog[2, 1/2 ((1 + I) + (1 - I) Tan[x/4])] +
PolyLog[2, 1/2 ((1 - I) + (1 + I) Tan[x/4])]))/(Log[
Cos[x/4]^2] + Log[1 - Tan[x/4]] + Log[1 + Tan[x/4]])

- Psymon

Whoops. Okay, it does converge, nvm x_x

- inkyvoyd

what am I not seeing sigh

- inkyvoyd

maybe u-sub?

- Psymon

Yeah, it converges, I just needed to do the actual work, lol.

- inkyvoyd

wait take x=cos theta

- inkyvoyd

wait no that doesn't work

- inkyvoyd

maybe we could use power series to reduce it into something more manageble?

- Psymon

You wanna see it?

- Psymon

I have an answer.

- inkyvoyd

hell yes

- Psymon

Now of course catch me if you see something I did wrong, haha.

- Psymon

Bleh, nvm, saw an error in what I was doing -_- lame.

- inkyvoyd

this is one of those problems that I don't think are solveable

- Psymon

Im too lazy to try trig sub, lol.

- inkyvoyd

no dude, I feel like it's not expressible in terms of elementary functions

- Psymon

Probably not. Ive gotten 50 different answers fromdifferent methods. And then none of them matched up with what he came up with. Meh, ohwell.

- inkyvoyd

dude, this problem is on magnitude of difficulty of the basel problem

- inkyvoyd

*I think

- inkyvoyd

how do you know this integral converges btw

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