解答本题时请勿使用图形计算器。 已知对于 $0 < x < \dfrac{\pi}{2}$，有 $\sin x > \dfrac{2x}{\pi}$。

1. Establish the inequality for the integrands.

Given that $\sin x > \frac{2x}{\pi}$ for $0 < x < \frac{\pi}{2}$, multiplying by $-1$ reverses the inequality sign:
$-\sin x < -\frac{2x}{\pi}$
Since $y = e^x$ is a strictly increasing function for all real $x$, applying the exponential function to both sides maintains the inequality:
$e^{-\sin x} < e^{-\frac{2x}{\pi}}$

2. Integrate both sides to conclude the proof.

Since $e^{-\sin x} < e^{-\frac{2x}{\pi}}$ for all $x$ in the interval $\left(0, \frac{\pi}{2}\right)$, the area under the curve $y = e^{-\sin x}$ is strictly less than the area under $y = e^{-\frac{2x}{\pi}}$ over this interval.

Integrating both sides with respect to $x$ from $0$ to $\frac{\pi}{2}$ gives:
$\int_{0}^{\frac{\pi}{2}} e^{-\sin x}\,dx < \int_{0}^{\frac{\pi}{2}} e^{-\frac{2x}{\pi}}\,dx$

1. Find the differential $dx$ in terms of $du$ and determine the new limits of integration.

Let $u = \pi - x$.
Then $\frac{du}{dx} = -1 \implies dx = -du$.

When $x = \frac{\pi}{2}$, $u = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.
When $x = \pi$, $u = \pi - \pi = 0$.

2. Substitute into the integral, apply trigonometric identities, and simplify.

$\int_{\frac{\pi}{2}}^{\pi} e^{-\sin x}\,dx = \int_{\frac{\pi}{2}}^{0} e^{-\sin(\pi - u)}\,(-du)$

Since $\sin(\pi - u) = \sin u$, this becomes:
$= -\int_{\frac{\pi}{2}}^{0} e^{-\sin u}\,du$

Reversing the limits of integration removes the negative sign:
$= \int_{0}^{\frac{\pi}{2}} e^{-\sin u}\,du$ (shown)

1. Split the integral from $0$ to $\pi$ and apply the result from part (ii).

$\int_{0}^{\pi} e^{-\sin x}\,dx = \int_{0}^{\frac{\pi}{2}} e^{-\sin x}\,dx + \int_{\frac{\pi}{2}}^{\pi} e^{-\sin x}\,dx$
From (ii), we know that $\int_{\frac{\pi}{2}}^{\pi} e^{-\sin x}\,dx = \int_{0}^{\frac{\pi}{2}} e^{-\sin u}\,du = \int_{0}^{\frac{\pi}{2}} e^{-\sin x}\,dx$.
$\therefore \int_{0}^{\pi} e^{-\sin x}\,dx = 2 \int_{0}^{\frac{\pi}{2}} e^{-\sin x}\,dx$

2. Apply the inequality from part (i).

From (i), $\int_{0}^{\frac{\pi}{2}} e^{-\sin x}\,dx < \int_{0}^{\frac{\pi}{2}} e^{-\frac{2x}{\pi}}\,dx$.
$\therefore \int_{0}^{\pi} e^{-\sin x}\,dx < 2 \int_{0}^{\frac{\pi}{2}} e^{-\frac{2x}{\pi}}\,dx$

3. Evaluate the definite integral of the exponential function.

$\int_{0}^{\frac{\pi}{2}} e^{-\frac{2x}{\pi}}\,dx = \left[ \frac{e^{-\frac{2x}{\pi}}}{-\frac{2}{\pi}} \right]_{0}^{\frac{\pi}{2}}$
$= -\frac{\pi}{2} \left( e^{-1} - e^0 \right)$
$= -\frac{\pi}{2} \left( \frac{1}{e} - 1 \right) = \frac{\pi}{2} \left( \frac{e-1}{e} \right)$

4. Substitute the evaluated integral back to complete the proof.

$2 \int_{0}^{\frac{\pi}{2}} e^{-\frac{2x}{\pi}}\,dx = 2 \left( \frac{\pi}{2} \right) \left( \frac{e-1}{e} \right) = \frac{\pi}{e}(e-1)$
$\therefore \int_{0}^{\pi} e^{-\sin x}\,dx < \frac{\pi}{e}(e-1)$