Finding the length of the curve y = ln(sin x), pi/6 <= x <= pi/2 the answer I come up with is ln(sqrt(3) + 1)

Question

does this look correct?

Answer 1

ds^2 = dx^2 + dy^2; from this you get the length as ∫[1/sinx]dx from π/6 to π/2; ∫[1/sinx}dx = ln[tan(x/2)]; evaluated at the limits this gives ln[tan(π/4)] - ln[tan(π/12)]

tan(π/4) = 1, so the first term is 0.

The second term is -ln[tan(π/12)] which comes out ln(2+√3)=1.317. This is not the same as your answer.

Find it worked out here http://img403.imageshack.us/img403/8256/segmentlengthel1.png

Answer 2

I believe your question is not accurate.
You are trying to find the lenght of a curve. To do so, yo need a starting and an end point.
In your question neither the starting or the ending point are fixed.
So a curve starting in pi/6 and ending in pi/2 is different to a curve going from pi/6.01 and pi/1.99.
Your question uses <= which makes impossible to be certain of the starting/ending point, this is the lenght of the curve.
The answer you are proposing is number which can be meausured, hence does not correspond to your question.

Answer 3

the equation for arc length is integral of sqrt(1+ f'(x)). let f(x) be y = ln(sinx) then f'(x) would be equal to 1/sinx * cosx = cosx/sinx = cotx. Therefore, you will be integrating sqrt (1+(cotx)^2) = sqrt (csc(x)^2) = csc(x)

Integrating csc(x) would give you ln(csc(x) - cot(x)) + C so integrating with lower limit pi/6 and upper limit pi/2 would give you the answer ln(csc(pi/2) - cot(pi/2)) - ln(csc(pi/6) - cot(pi/6)) = ln(1) - ln(2 - sqrt(3)) = - ln(2 - sqrt(3))

Therefore, the answer is - ln(2 - sqrt(3)) which is equal to 1.317

Answer 4

the frmula is sqrt(1+(f(x)')^2) so the answer is the integral of sqrt(1+(cotg(x))^2)= integral of 1/sin x now put x=2arctg(t) and proceed.