find the length of x=ln cos y from y= -pie/4 to y= pie/4
The equation to use, for distance along a line, is
s = int [y1 to y2] √(1 + (x’)^2) dy
We have x = ln(cos(y))
x’ = -sin(y)/cos(y)
1 + (x’)^2 = 1 + sin^2(y)/cos^2(y) = (cos^2(y) + sin^2(y))/cos^2(y) = 1/cos^2(y)
√(1 + (x’)^2) = 1/cos(y) = sec(y)
Hence,
s = int [y1=-π/4 to y2=π/4] sec(y) dy
s = [ln(sec(y) + tan(y)] [-π/4 to π/4]
s = {ln(sec(π/4] + tan(π/4) – ln(sec(-π/4) + tan(-π/4)}
s = {ln(√2 + 1) – ln(√2 – 1)}
s = ln{(√2 + 1) / (√2 – 1)}
s = ln{(√2 + 1)^2 / (√2 – 1)(√2 + 1)} = ln{(2 + 2√2 + 1)^2 / (2 – 1)}
s = ln(2√2 + 3)