Finding the Arc Length of a Graph

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.

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*}