need help in finding the length of the curve

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need help in finding the length of the curve

Mathematics
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y = 2ln(sin(x/2) pi/3
i got it down to the integral of csc(x/2)

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can you show your working to get to that point?
yeah i'll try to draw it out
|dw:1443753630087:dw|
there
sin^2 + cos^2 = 1 1 + cos^2/sin^2 = csc^2 ok, im on the same page as you now
yeah thats what i did for the derivative
an old trick is to multiply by (csc + cot) i beleive
Oh cosecant integral :) This is one of those annoying ones. Do you remember secant integral? it's very similar process for this one
yeah i got the integral for cosecant its -ln|cscx+cotx|
csc + cot derives to -(csc^2 + csc cot)
the x/2 pops out tho so we need to adjust by a constant right?
i guess
-ln(cscu + cotu) (csc^2u + cscu cotu) u' -------------------- csc u + cot u u = x/2; u' = 1/2
-2 ln ... seems fair to me
wait how did you get csc^2u + cscu cotu?
what is the derivative of csc u + cot u ?
oh but shouldnt it be negative
maybe ...
why are we taking the derivative of those two if we already got the integral?
|dw:1443754334011:dw|
what is your question then?
my question is why is my answer different from that of wolfram-alpha i got -2ln|1-2sqrt(3)|
and they got -2ln|2-sqrt(3)|?
i dont know if there is a step you gotta do before plugging in the integration numbers or what?
\[-2ln(csc(pi/2)+cot(pi/2)+2ln(csc(pi/6)+cot(pi/6))\] \[2[-ln(csc(pi/2)+cot(pi/2)+ln(csc(pi/6)+cot(pi/6))\] \[2ln(\frac{csc(pi/6)+cot(pi/6)}{csc(pi/2)+cot(pi/2)})\]
what the....?
okay so csc(pi/6) is 2 and cot(pi/6) is sqrt(3)
and csc(pi/2) is 1 and cot(pi/2) is 0
so |dw:1443755083983:dw|
but dont you do that only when your substracting?
??
when do you divide the numbers when there is a natural log i mean?
-2 ln|2-sqrt(3)| 2 ln|1/(2-sqrt(3))| 2 ln|(2+sqrt(3))/(4-3)|
log(a) - log(b) = log(a/b) for all logs
but your adding them?
no, -2ln(b) --2ln(a) = 2ln(a)-2ln(b)
= 2 ln(a/b)
hey where you guys at... have you guys gotten to \[2 \int\limits_\frac{\pi}{6}^\frac{\pi}{2} \csc(u) du\] if this looks funny I just subbed x/2 for u
i think you have the b on top of the a
i was trapped in OS purgatory ...

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