Answer
the length of the sine curve:
$$
y= \sin x, \quad \quad 0 \leq x \leq \pi
$$
is given by
$$
\begin{aligned}
L &=\int_{0}^{\pi}f(x ) dx \\
&=\int_{0}^{\pi} \sqrt{1+\left(\cos x\right)^{2}} d x \\
& \approx 3.820188 \quad\quad\left[\text {by using Simpson’s Rule } \right] \\
\end{aligned}
$$
Work Step by Step
$$
y= \sin x, \quad \quad 0 \leq x \leq \pi
$$
$\Rightarrow$
$$
y^{\prime} =\frac{dy}{dx}=\cos x
$$
$\Rightarrow$
$$
1+\left(\frac{d y}{d x}\right)^{2} =1+\left(\cos x\right)^{2} .
$$
Let
$$
f(x)=\sqrt{1+\left(\frac{d y}{d x}\right)^{2}} =\sqrt{1+\left(\cos x\right)^{2}}
$$
Then the arc length is
$$
\begin{aligned}
L &=\int_{0}^{\pi}f(x ) dx \\
&=\int_{0}^{\pi}\sqrt{1+\left(\frac{d y}{d x}\right)^{2}} dx\\
&=\int_{0}^{\pi} \sqrt{1+\left(\cos x\right)^{2}} d x \\
\end{aligned}
$$
Use Simpson’s Rule to estimate the length of the sine curve with with $ n =10 , a = 0$, and $b = \pi$:
$$
\Delta x=\frac{b-a}{n}=\frac{\pi-0}{10}=\frac{\pi}{10}
$$ gives :
$$
\begin{aligned} \int_{0}^{\pi} \sqrt{1+\left(\cos x\right)^{2}} d x & \approx \\
& \approx \frac{\Delta x}{3}\left[f\left(x_{0}\right)\right. +4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)+\cdots \\
& \quad \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right] \\
&= S_{10} \\
& = \frac{(\pi-0) / 10}{3}\left[f(0)+4 f\left(\frac{\pi}{10}\right)+\\
+2 f\left(\frac{2 \pi}{10}\right)+4 f\left(\frac{3 \pi}{10}\right)+2 f\left(\frac{4 \pi}{10}\right)\right.\\ &\left.+4 f\left(\frac{5 \pi}{10}\right) +2 f\left(\frac{6 \pi}{10}\right)+4 f\left(\frac{7 \pi}{10}\right)+\\ +2 f\left(\frac{8 \pi}{10}\right)+4 f\left(\frac{9 \pi}{10}\right)+f(\pi)\right]\\
&\approx 3.820188 \end{aligned}
$$
