Posted by sammyg on Tuesday, January 24, 2012 at 9:46pm.
y = ln sinx
y' = 1/sinx * cosx = tanx
s = Int(sqrt(1+(y')^2)dx)[pi/4,3pi/4]
= Int(sqrt(1+tan^2(x))dx)[pi/4,3pi/4=
= Int(secx dx)[pi/4,3pi/4]
= ln|secx + tanx|[pi/4,3pi/4]
= ln|-1/√2 + 1| - ln|1/√2 + 1|
= ln|(1-√2/(1+√2)|
= ln|2√2-3|
= ln(3-2√2)
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