Q. 44 - Calculus

Question

Find the exact value of the arc length of each function f(x) on [a, b] by writing the arc length as a definite integral and then solving that integral.

$f\left(x\right)=\ln \left(\cos x\right)$, $\left[a,b\right]=\left[0,\frac{\pi }{4}\right]$

Step-by-Step Solution

Answer

The arc length is $\begin{array}{rcl}& & \ln \left(1+\sqrt{2}\right)\end{array}$.

Step 1. Given information.

Consider the function is $f\left(x\right)=\ln \left(\cos x\right)$, $\left[a,b\right]=\left[0,\frac{\mathrm{\pi }}{4}\right]$.

Step 2. Use arc length formula.

The formula for a function to find the arc length from $x=a$ to $x=b$ is given by ${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Step 3. Find the arc length.

$\begin{array}{rcl}\mathrm{Arc} \mathrm{length}& =& {\int }_0^{\mathrm{\pi }/4}\sqrt{1+{\left(\frac{d}{dx}\left(\ln \left(\cos x\right)\right)\right)}^2}dx\\ & =& {\int }_0^{\mathrm{\pi }/4}\sqrt{1+{\left(\frac{1}{\cos x}\left(-\sin x\right)\right)}^2}dx\\ & =& {\int }_0^{\mathrm{\pi }/4}\sqrt{1+{\left(-\tan x\right)}^2}dx\\ & =& {\int }_0^{\mathrm{\pi }/4}\sqrt{1+{\tan }^{2}x}dx\\ & =& {\int }_0^{\mathrm{\pi }/4}secxdx\\ & =& {\left[\ln \left|\tan \left(x\right)+sec\left(x\right)\right|\right]}_0^{\mathrm{\pi }/4}\\ & =& \left[\ln \left|\tan \left(\frac{\mathrm{\pi }}{4}\right)+sec\left(\frac{\mathrm{\pi }}{4}\right)\right|\right]-\ln \left|\tan \left(0\right)+sec\left(0\right)\right|\\ & =& \ln \left(1+\sqrt{2}\right)\\ & & \end{array}$