#### Answer

For $N=10$:
The midpoint approximation of the length $\simeq 6.90373$.
For $N=20$:
The midpoint approximation of the length $\simeq 6.91504$.
For $N=30$:
The midpoint approximation of the length $\simeq 6.91495$.
For $N=50$:
The midpoint approximation of the length $\simeq 6.91495$.

#### Work Step by Step

We have
$x\left( t \right) = \cos t$, ${\ \ \ }$ $x'\left( t \right) = - \sin t$,
$y\left( t \right) = {{\rm{e}}^{\sin t}}$, ${\ \ \ }$ $y'\left( t \right) = {{\rm{e}}^{\sin t}}\cos t$.
Using Eq. (3) of Theorem 1, the arc length for $0 \le t \le 2\pi $ is
$s = \mathop \smallint \limits_0^{2\pi } \sqrt {{{\left( { - \sin t} \right)}^2} + {{\left( {{{\rm{e}}^{\sin t}}\cos t} \right)}^2}} {\rm{d}}t$
To approximate the definite integral above using Midpoint Rule, we divide $\left[ {0,2\pi } \right]$ into $N$ subintervals.
The length of the interval is $\Delta x = \frac{{2\pi - 0}}{N} = \frac{{2\pi }}{N}$.
The midpoints are ${c_j} = \left( {j - \frac{1}{2}} \right)\Delta x$ for $j = 1,2,...,N$.
The midpoint approximation becomes
${M_N} = \frac{{2\pi }}{N}\mathop \sum \limits_{j = 1}^N f\left( {{c_j}} \right)$,
where $f\left( x \right) = \sqrt {{{\left( { - \sin x} \right)}^2} + {{\left( {{{\rm{e}}^{\sin x}}\cos x} \right)}^2}} $.
For $N=10$:
We have $\Delta x = \frac{{2\pi }}{{10}} = \frac{\pi }{5}$ and ${c_j} = \frac{\pi }{5}\left( {j - \frac{1}{2}} \right)$, for $j=1,2,...,10$.
The midpoint approximation is
${M_{10}} = \frac{\pi }{5}\mathop \sum \limits_{j = 1}^{10} f\left( {{c_j}} \right) \simeq 6.90373$.
For $N=20$:
We have $\Delta x = \frac{{2\pi }}{{20}} = \frac{\pi }{{10}}$ and ${c_j} = \frac{\pi }{{10}}\left( {j - \frac{1}{2}} \right)$, for $j=1,2,...,20$.
The midpoint approximation is
${M_{20}} = \frac{\pi }{{10}}\mathop \sum \limits_{j = 1}^{20} f\left( {{c_j}} \right) \simeq 6.91504$.
For $N=30$:
We have $\Delta x = \frac{{2\pi }}{{30}} = \frac{\pi }{{15}}$ and ${c_j} = \frac{\pi }{{15}}\left( {j - \frac{1}{2}} \right)$, for $j=1,2,...,30$.
The midpoint approximation is
${M_{30}} = \frac{\pi }{{15}}\mathop \sum \limits_{j = 1}^{30} f\left( {{c_j}} \right) \simeq 6.91495$.
For $N=50$:
We have $\Delta x = \frac{{2\pi }}{{50}} = \frac{\pi }{{25}}$ and ${c_j} = \frac{\pi }{{25}}\left( {j - \frac{1}{2}} \right)$, for $j=1,2,...,50$.
The midpoint approximation is
${M_{50}} = \frac{\pi }{{25}}\mathop \sum \limits_{j = 1}^{50} f\left( {{c_j}} \right) \simeq 6.91495$.