Q. 45 - 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.

$f\left(x\right)=\ln \left(\sin x\right)$,  $\left[a,b\right]=\left[\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}\right]$

Step-by-Step Solution

Answer

The arc length is $\ln \left(3+2\sqrt{2}\right)$.

Step 1. Given information.

Consider the function is $f\left(x\right)=\ln \left(\sin x\right)$, $\left[a,b\right]=\left[\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}\right]$.

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.

$\begin{array}{rcl}\mathrm{Arc} \mathrm{length}& =& {\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4}\sqrt{1+{\left(\frac{d}{dx}\left(\ln \left(\sin x\right)\right)\right)}^2}dx\\ & =& {\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4}\sqrt{1+{\left(\frac{1}{\sin x}\left(\cos x\right)\right)}^2}dx\\ & =& {\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4}\sqrt{1+{\left(cotx\right)}^2}dx\\ & =& {\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4}\sqrt{1+co{t}^{2}x}dx\\ & =& {\int }_{\mathrm{\pi }/4}^{3\mathrm{\pi }/4}\cos ecxdx\\ & =& {\int }_{\sqrt{2}-1}^{\sqrt{2}-1}\frac{1}{u}du\\ & =& {\left[\ln \left|u\right|\right]}_{\sqrt{2}-1}^{\sqrt{2}+1}\\ & =& \ln \left|\sqrt{2}+1\right|-\ln \left|\sqrt{2}-1\right|\\ & =& \ln \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\\ & =& \ln \left(3+2\sqrt{2}\right)\\ & & \end{array}$