Solve the problem.
-Find the length of the curve $y = \ln ( \sin x ) , \pi / 3 \leq x \leq \pi / 2$
A) $\ln ( \sqrt { 3 } + 1 )$
B) $1 - \ln ( \sqrt { 3 } )$
C) $\ln ( \sqrt { 3 } )$
D) $\ln ( 2 \sqrt { 3 } )$

Question

Quizplus · Accepted Answer

The formula for the length of a curve $y=f(x)$ from $x=a$ to $x=b$ is 
$$\int_{a}^{b} \sqrt{1+f'(x)^2} dx$$
Here, we have $f(x) = \ln(\sin x)$, so $f'(x) = \frac{\cos x}{\sin x}$. Thus, 
\begin{align*}
\sqrt{1+f'(x)^2} &= \sqrt{1+\frac{\cos^2 x}{\sin^2 x}} \
&= \sqrt{\frac{\sin^2 x + \cos^2 x}{\sin^2 x}} \
&= \frac{1}{|\sin x|}
\end{align*}
since $\sin x \leq 0$ on the given interval. Thus, the length of the curve is 
$$\int_{\pi/3}^{\pi/2} \frac{1}{|\sin x|} dx.$$
We can evaluate this integral using the substitution $u = \cos x$, so $du = -\sin x dx$. Then 
\begin{align*}
\int_{\pi/3}^{\pi/2} \frac{1}{|\sin x|} dx &= \int_{1/2}^{1} \frac{1}{\sqrt{1-u^2}} du \
&= \arcsin(u) \bigg|_{1/2}^1 \
&= \arcsin(1) - \arcsin(1/2) \
&= \frac{\pi}{2} - \frac{\pi}{6} \
&= \frac{\pi}{3}.
\end{align*}
Finally, we have 
$$\ln(\sqrt{3}) < \ln(\sin x) < 0$$
on the given interval, so 
$$\ln(\sqrt{3}) < y < 0.$$
Thus, the answer is $\boxed{\textbf{(C)}}$.

Solve the Problem y=ln⁡(sin⁡x),π/3≤x≤π/2y = \ln ( \sin x ) , \pi / 3 \leq x \leq \pi / 2y=ln(sinx),π/3≤x≤π/2

Solve the Problem $y = \ln ( \sin x ) , \pi / 3 \leq x \leq \pi / 2$