calculus
posted by sammyg .
Find the arc length of the graph of the function over the indicated interval. (Round your answer to three decimal places.)
y = ln (sin(x))
,[π/4,3π/4]

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]
= lnsecx + tanx[pi/4,3pi/4]
= ln1/√2 + 1  ln1/√2 + 1
= ln(1√2/(1+√2)
= ln2√23
= ln(32√2)
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