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{\bf Exam Question

Topic: Arc Length}

Calculate the length of the curve given by $$y=\ln(\cos x);\
(-\frac{\pi}{4}\le x\le \frac{\pi}{4}).$$ Give your answer both in
exact form and also as an approximation rounded to four decimal
places, using your calculator. \vspace{0.5in}

{\bf Solution}


$$L=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{1+f'(x)^2}\, dx,$$
where $f(x)=\ln(\cos x)$

Now $f'(x)=\tan x\ \mathrm{and}\ 1+\tan^2x=\sec^2x.$


\begin{eqnarray*} L & = & \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec
x\, dx=\left[\ln(\sec x+\tan
x)\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\\
&=&\ln(\sqrt2+1)-\ln(\sqrt2-1)=\ln\left(\frac{\sqrt2+1}{\sqrt2-1}\right)\\
&=&\ln(\sqrt2+1)^2=\ln(3+2\sqrt2)=1.7627 \mathrm{(4\ d.p.)}
\end{eqnarray*}








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