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We start by simplifying the expression inside the integral. We have $$ \frac{1-\sin x}{1-\cos x} = \frac{1-\sin x}{1-\cos x} \cdot \frac{1+\cos x}{1+\cos x} = \frac{(1-\sin x)(1+\cos x)}{1-\cos^2 x} = \frac{1+\cos x - \sin x - \sin x \cos x}{\sin^2 x} $$ $$ = \frac{1}{\sin^2 x} + \frac{\cos x}{\sin^2 x} - \frac{\sin x}{\sin^2 x} - \frac{\sin x \cos x}{\sin^2 x} = \csc^2 x + \frac{\cos x}{\sin^2 x} - \frac{1}{\sin x} - \frac{\cos x}{\sin x} $$ $$ = \csc^2 x + \csc x \cot x - \csc x - \cot x $$ However, a more useful simplification is: $$ \frac{1-\sin x}{1-\cos x} = \frac{1-2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} = \frac{\sin^2(x/2) + \cos^2(x/2) - 2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} $$ $$ = \frac{(\cos(x/2) - \sin(x/2))^2}{2\sin^2(x/2)} = \frac{1}{2} \left( \frac{\cos(x/2) - \sin(x/2)}{\sin(x/2)} \right)^2 = \frac{1}{2} \left( \cot(x/2) - 1 \right)^2 $$ $$ = \frac{1}{2} (\cot^2(x/2) - 2\cot(x/2) + 1) = \frac{1}{2} (\csc^2(x/2) - 1 - 2\cot(x/2) + 1) = \frac{1}{2} \csc^2(x/2) - \cot(x/2) $$ Thus, the integral becomes $$ \int_{\pi/2}^{\pi} e^x \left( \frac{1}{2} \csc^2(x/2) - \cot(x/2) \right) dx $$
Let $I = \int_{\pi/2}^{\pi} e^x \left( \frac{1}{2} \csc^2(x/2) - \cot(x/2) \right) dx$. We can rewrite this as $I = \int_{\pi/2}^{\pi} e^x \left( -\cot(x/2) + \frac{1}{2} \csc^2(x/2) \right) dx$. Consider the function $f(x) = -\cot(x/2)$. Then $f'(x) = \frac{1}{2} \csc^2(x/2)$. Thus, we have $I = \int_{\pi/2}^{\pi} e^x (f(x) + f'(x)) dx$. We know that $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$. Therefore, $I = \left[ e^x (-\cot(x/2)) \right]_{\pi/2}^{\pi} = - \left[ e^x \cot(x/2) \right]_{\pi/2}^{\pi}$.
Now we evaluate the limits: $$ I = - \left[ e^{\pi} \cot(\pi/2) - e^{\pi/2} \cot(\pi/4) \right] = - \left[ e^{\pi} (0) - e^{\pi/2} (1) \right] = - \left[ 0 - e^{\pi/2} \right] = e^{\pi/2} $$
Final Answer: $e^{\pi/2}$
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