Chapter 8 - Further Applications of Integration - 8.1 Arc Length - 8.1 Exercises - Page 589: 16

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

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