Answer
$\displaystyle \frac{6}{5}$
Work Step by Step
$L=\displaystyle \int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^{2}}dx$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[\frac{1}{2}\cosh 2x]=\frac{1}{2}\cdot\sinh 2x\cdot 2=\sinh 2x$
$L=\displaystyle \int_{0}^{\ln\sqrt{5}}\sqrt{1+\sinh^{2}2x}dx$
Apply the identity: $\cosh^{2}x-\sinh^{2}x=1$
$L=\displaystyle \int_{0}^{\ln\sqrt{5}}\cosh 2xdx=\frac{1}{2}[\sinh 2x]_{0}^{\ln\sqrt{5}}$
Evaluate:
$\displaystyle \sinh(2\ln\sqrt{5})=\sinh(2\cdot\frac{1}{2}\ln 5)=\sinh(5)$
$=\displaystyle \frac{e^{\ln 5}-e^{-\ln 5}}{2}=\frac{5-\frac{1}{5}}{2}\cdot\frac{5}{5}=\frac{25-1}{10}=\frac{12}{5}$
$\displaystyle \sinh 0=\frac{1-1}{1+1}=0$
$L=\displaystyle \frac{1}{2}\cdot\frac{12}{5}=\frac{6}{5}$
