Chapter 8 - Section 8.1 - Arc Length - 8.1 Exercises - Page 549: 16

$\frac{1}{2} \sinh 2$

Given: $y = 3+ \frac{1}{2} \cosh 2x$ $y' = \sinh 2x$ and $1+(dy/dx)^{2} = 1 + \sinh^{2} 2x = \cosh^{2} 2x$ $L = \int^{1}_{0} \sqrt{\cosh^{2} 2x} dx $ $= \int^{1}_{0} \cosh 2x dx $ $= [\frac{1}{2} \sinh 2x]^{1}_{0}$ $ = \frac{1}{2} \sinh 2 - 0 $ $=\frac{1}{2} \sinh 2$