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

Question

Quizplus · Accepted Answer

The length $L$ of a curve defined by $y = f(x)$ from $x = a$ to $x = b$ is given by the integral $L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$. For $y = \ln(\csc(x))$, we first find $\frac{dy}{dx}$. $\frac{dy}{dx} = \frac{d}{dx} \ln(\csc(x)) = \frac{d}{dx} \ln(\frac{1}{\sin(x)}) = -\cot(x)$.Thus, $\left(\frac{dy}{dx}\right)^2 = \cot^2(x)$.The integral becomes $L = \int_{\pi/4}^{\pi/2} \sqrt{1 + \cot^2(x)} dx$.Since $1 + \cot^2(x) = \csc^2(x)$, the integral simplifies to $L = \int_{\pi/4}^{\pi/2} |\csc(x)| dx$.Given the interval $\pi/4 \leq x \leq \pi/2$, $\csc(x)$ is positive, so we can drop the absolute value, leading to $L = \int_{\pi/4}^{\pi/2} \csc(x) dx$.The antiderivative of $\csc(x)$ is $-\ln|\csc(x) + \cot(x)|$, so we evaluate this from $\pi/4$ to $\pi/2$:$L = -\ln|\csc(\pi/2) + \cot(\pi/2)| + \ln|\csc(\pi/4) + \cot(\pi/4)|$.Since $\csc(\pi/2) = 1$, $\cot(\pi/2) = 0$, $\csc(\pi/4) = \sqrt{2}$, and $\cot(\pi/4) = 1$, we get:$L = -\ln|1 + 0| + \ln|\sqrt{2} + 1|$,which simplifies to $L = \ln(\sqrt{2} + 1)$.Therefore, the correct answer is A) $\ln(\sqrt{2} + 1)$.

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

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