Find the arc length of the graph of the function $f(x) =\ln~\sec~x$ from $x=0$ to $x=\dfrac{\pi}{4}$.

Solution The graph of $f(x) =\ln~\sec~x$ is shown in Figure 49.

Figure 49 $f(x) =\ln~\sec~x, 0\leq x\leq \dfrac{\pi}{4}$.

The derivative of $f$ is $f' (x)=\dfrac{\sec x\tan x}{\sec x}=\tan x$. Since $f' (x) =\tan x$ is continuous on the open interval $\left( -\dfrac{\pi}{2},\dfrac{\pi}{2} \right) ,$ which contains $0$ and $\dfrac{\pi}{4},$ we use the arc length formula (1). The arc length $s$ from $0$ to $\dfrac{\pi}{4}$ is $$\begin{align*} s&=&\int_{0}^{\pi /4}\sqrt{~1+[ f' (x) ] ^{2}}~ {\it dx}=\int_{0}^{\pi /4}\sqrt{1+\tan ^{2}x}\,{\it dx}=\int_{0}^{\pi /4}\sqrt{\sec ^{2}x }\,{\it dx} \\[5pt] &=&\int_{0}^{\pi /4}\sec x\,{\it dx}=\big[ \ln \vert \sec x+\tan x\vert \big] _{0}^{\pi /4} \\[5pt] &=&\ln \left\vert \sec \dfrac{\pi}{4}+\tan \dfrac{\pi}{4}\right\vert -\ln \left\vert \sec 0+\tan 0\right\vert =\ln \vert \sqrt{2}+1\vert \end{align*}$$