Solve the problem.
-Find the length of the curve $y = \ln ( \sin x ) , \pi / 4 \leq x \leq \pi / 2$
A) $1 - \ln ( \sqrt { 2 } + 1 )$
B) $\ln ( 2 \sqrt { 2 } )$
C) $\ln ( \sqrt { 2 } )$
D) $\ln ( \sqrt { 2 } + 1 )$

Question

Quizplus · Accepted Answer

Using the formula for arc length of a curve, we have
$$L = \int_{\pi/4}^{\pi/2} \sqrt{1 + (\frac{d}{dx}\ln(\sin x))^2} dx$$
$$= \int_{\pi/4}^{\pi/2} \sqrt{1 + \frac{(\cos x)^2}{(\sin x)^2}} dx$$
$$= \int_{\pi/4}^{\pi/2} \sqrt{\frac{(\sin x)^2+(\cos x)^2}{(\sin x)^2}} dx$$
$$= \int_{\pi/4}^{\pi/2} \sqrt{\frac{1}{(\sin x)^2}} dx$$
$$= \int_{\pi/4}^{\pi/2} \csc x dx$$
Using integration by substitution, $u = \cot x$, $du = -\csc^2 x \ dx$, we have
$$L = -\int_{\cot (\pi/4)}^{\cot (\pi/2)} du = -[\ln(\cot x)]_{\cot (\pi/4)}^{\cot (\pi/2)} = \ln (\sqrt{2} + 1)$$
Therefore, the answer is D.

Solve the Problem y=ln⁡(sin⁡x),π/4≤x≤π/2y = \ln ( \sin x ) , \pi / 4 \leq x \leq \pi / 2y=ln(sinx),π/4≤x≤π/2

Solve the Problem $y = \ln ( \sin x ) , \pi / 4 \leq x \leq \pi / 2$