Question

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

Answer 1

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

Answer 2

the first one

Answer 3

I am sorry, I really don't know,

Answer 4

thats ok neither do i

Answer 5

this is very complicated stuff

Answer 6

@ash2326

Answer 7

\[\frac{ 4 }{ \sin(x)\sqrt{1-x ^{2}} }\]?

What loser tried to post shows as an error for me, so just want to confirm if this is it

Answer 8

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

Answer 9

maybe integration by parts and trig sub?

Answer 10

how so?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

ok i'll try that thanks

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

does that integral even converge btw?>

Answer 21

Nope.

Answer 22

LOL

Answer 23

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

Answer 24

yeah guy42 use limits

Answer 25

it doesn't converge? are you sure?

Answer 26

Positive.

Answer 27

ok thanks

Answer 28

nope it does converge

Answer 29

gg

Answer 30

Really? O.o

Answer 31

the not converge answer choice was wrong

Answer 32

do you get another try?

Answer 33

3 more tries

Answer 34

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

Answer 35

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

Answer 36

wolfram goes all dead trying to do it

Answer 37

kay
lemme try mathematica

Answer 38

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

Answer 39

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

Answer 40

i got the answer (pi^2)/2

Answer 41

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

Answer 42

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

Answer 43

Yeah, I did.

Answer 44

does it work?

Answer 45

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

Answer 46

ok guys im gonna go thanks for your help

Answer 47

e

Answer 48

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]])

Answer 49

Whoops. Okay, it does converge, nvm x_x

Answer 50

what am I not seeing sigh

Answer 51

maybe u-sub?

Answer 52

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

Answer 53

wait take x=cos theta

Answer 54

wait no that doesn't work

Answer 55

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

Answer 56

You wanna see it?

Answer 57

I have an answer.

Answer 58

hell yes

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

Im too lazy to try trig sub, lol.

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

*I think

Answer 67

how do you know this integral converges btw