Evaluate the definite integral $\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$.

(D) Let $I = \int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$.
Using trigonometric identities $1-\sin x = 1-2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1-\cos x = 2 \sin^{2} \frac{x}{2}$,we get:
$I = \int_{\frac{\pi}{2}}^{\pi} e^{x} \left( \frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^{2} \frac{x}{2}} \right) d x$
$I = \int_{\frac{\pi}{2}}^{\pi} e^{x} \left( \frac{1}{2} \csc^{2} \frac{x}{2} - \cot \frac{x}{2} \right) d x$
Let $f(x) = -\cot \frac{x}{2}$. Then $f'(x) = -(-\frac{1}{2} \csc^{2} \frac{x}{2}) = \frac{1}{2} \csc^{2} \frac{x}{2}$.
Since $\int e^{x} (f(x) + f'(x)) d x = e^{x} f(x) + C$,we have:
$I = \left[ e^{x} (-\cot \frac{x}{2}) \right]_{\frac{\pi}{2}}^{\pi}$
$I = -\left[ e^{\pi} \cot \frac{\pi}{2} - e^{\frac{\pi}{2}} \cot \frac{\pi}{4} \right]$
Since $\cot \frac{\pi}{2} = 0$ and $\cot \frac{\pi}{4} = 1$:
$I = -\left[ e^{\pi} (0) - e^{\frac{\pi}{2}} (1) \right] = e^{\frac{\pi}{2}}$.