anonymous
  • anonymous
determine if the improper integral of 4/sin(x)(sqrt(1-x^2) from 0 to 1converges and if so at what value?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Loser66
  • 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
  • anonymous
the first one
Loser66
  • Loser66
I am sorry, I really don't know,

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More answers

anonymous
  • anonymous
thats ok neither do i
anonymous
  • anonymous
this is very complicated stuff
Loser66
  • Loser66
@ash2326
Psymon
  • 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
  • anonymous
yes also you take the integral of that from 0 to 1
inkyvoyd
  • inkyvoyd
maybe integration by parts and trig sub?
anonymous
  • anonymous
how so?
inkyvoyd
  • inkyvoyd
well it looks like your u and v' will be 1/sin x and 1/sqrt(1-x^2)
inkyvoyd
  • 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
  • Psymon
Yeah, that does work out actually. I guess I was just double checking myself.
inkyvoyd
  • inkyvoyd
btw that's my excuse for not doing the problem since I don't feel like it lol
anonymous
  • anonymous
ok i'll try that thanks
Psymon
  • Psymon
Mhm. You'll have to do integration by parts twice. It comes out pretty clean, though.
inkyvoyd
  • inkyvoyd
protip if you're allowed to use a table for the trig sub, it's a pain in the arse
inkyvoyd
  • inkyvoyd
@Psymon want a even simpler integration problem that's really hard to evaluate?
Psymon
  • Psymon
Lol, why would I want to do more unnecessary work? xD This one just turned out to be okay.
inkyvoyd
  • inkyvoyd
does that integral even converge btw?>
Psymon
  • Psymon
Nope.
inkyvoyd
  • inkyvoyd
LOL
Psymon
  • Psymon
But just saying that isnt enough, he still has to be able to work his way through it to determine that, lol.
inkyvoyd
  • inkyvoyd
yeah guy42 use limits
anonymous
  • anonymous
it doesn't converge? are you sure?
Psymon
  • Psymon
Positive.
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
nope it does converge
inkyvoyd
  • inkyvoyd
gg
Psymon
  • Psymon
Really? O.o
anonymous
  • anonymous
the not converge answer choice was wrong
inkyvoyd
  • inkyvoyd
do you get another try?
anonymous
  • anonymous
3 more tries
Psymon
  • 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
  • inkyvoyd
psymon go to wolfram alpha, and someone gimme a medal or inky be sad
Psymon
  • Psymon
wolfram goes all dead trying to do it
inkyvoyd
  • inkyvoyd
kay lemme try mathematica
Psymon
  • 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
  • Psymon
Oh wow.....I think I got an answer x_x This is one funky integral, questionable accuracy like crazy.
anonymous
  • anonymous
i got the answer (pi^2)/2
Psymon
  • Psymon
And thats still not the answer I got, haha. Like seriously, wtf?
inkyvoyd
  • inkyvoyd
ok, did you guys try out what I suggested with the by parts and all?
Psymon
  • Psymon
Yeah, I did.
inkyvoyd
  • inkyvoyd
does it work?
Psymon
  • 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
  • anonymous
ok guys im gonna go thanks for your help
Psymon
  • Psymon
e
inkyvoyd
  • 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
  • Psymon
Whoops. Okay, it does converge, nvm x_x
inkyvoyd
  • inkyvoyd
what am I not seeing sigh
inkyvoyd
  • inkyvoyd
maybe u-sub?
Psymon
  • Psymon
Yeah, it converges, I just needed to do the actual work, lol.
inkyvoyd
  • inkyvoyd
wait take x=cos theta
inkyvoyd
  • inkyvoyd
wait no that doesn't work
inkyvoyd
  • inkyvoyd
maybe we could use power series to reduce it into something more manageble?
Psymon
  • Psymon
You wanna see it?
Psymon
  • Psymon
I have an answer.
inkyvoyd
  • inkyvoyd
hell yes
Psymon
  • Psymon
Now of course catch me if you see something I did wrong, haha.
Psymon
  • Psymon
Bleh, nvm, saw an error in what I was doing -_- lame.
inkyvoyd
  • inkyvoyd
this is one of those problems that I don't think are solveable
Psymon
  • Psymon
Im too lazy to try trig sub, lol.
inkyvoyd
  • inkyvoyd
no dude, I feel like it's not expressible in terms of elementary functions
Psymon
  • 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
  • inkyvoyd
dude, this problem is on magnitude of difficulty of the basel problem
inkyvoyd
  • inkyvoyd
*I think
inkyvoyd
  • inkyvoyd
how do you know this integral converges btw

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