**Step 1: Simplify the integrand using trigonometric identities** We have: $\frac{1 - \sin 2x}{1 - \cos 2x} = \frac{1 - 2\sin x \cos x}{2\sin^2 x} = \frac{1}{2\sin^2 x} - \frac{2\sin x \cos x}{2\sin^2 x} = \frac{1}{2} \csc^2 x - \cot x$

**Step 2: Rewrite the integral** $\int_{\pi/4}^{\pi/2} e^{2x} (\frac{1 - \sin 2x}{1 - \cos 2x}) dx = \int_{\pi/4}^{\pi/2} e^{2x} (\frac{1}{2} \csc^2 x - \cot x) dx$

**Step 3: Manipulate the integrand to match the form $\int e^{ax} [f(x) + f'(x)] dx = e^{ax} f(x) + C$** We want to express the integrand in the form $e^{2x} [f(x) + f'(x)]$. Let's try to rewrite the integral as: $\int_{\pi/4}^{\pi/2} e^{2x} (-\cot x + \frac{1}{2} \csc^2 x) dx = -\frac{1}{2} \int_{\pi/4}^{\pi/2} e^{2x} (2\cot x - \csc^2 x) dx$ Now, consider $f(x) = \cot x$. Then $f'(x) = -\csc^2 x$. We need to adjust the integral to fit the form $\int e^{ax} [f(x) + \frac{1}{a}f'(x)] dx = e^{ax} \frac{f(x)}{a} + C$. Let's rewrite the integral as: $\int_{\pi/4}^{\pi/2} e^{2x} (-\cot x + \frac{1}{2} \csc^2 x) dx = \int_{\pi/4}^{\pi/2} e^{2x} (-\cot x + \frac{1}{2} \csc^2 x) dx$ We can rewrite this as: $\int_{\pi/4}^{\pi/2} e^{2x} (-\cot x + \frac{1}{2} \csc^2 x) dx = \int_{\pi/4}^{\pi/2} e^{2x} (-\cot x + \frac{1}{2} (-\cot x)') dx$ Using the formula $\int e^{ax} [f(x) + \frac{1}{a}f'(x)] dx = \frac{e^{ax}f(x)}{a} + C$, we have: $\int_{\pi/4}^{\pi/2} e^{2x} (-\cot x + \frac{1}{2} \csc^2 x) dx = \left[ e^{2x} \frac{-\cot x}{2} \right]_{\pi/4}^{\pi/2}$

**Step 4: Evaluate the definite integral** $\left[ -\frac{1}{2} e^{2x} \cot x \right]_{\pi/4}^{\pi/2} = -\frac{1}{2} \left[ e^{2(\pi/2)} \cot(\pi/2) - e^{2(\pi/4)} \cot(\pi/4) \right] = -\frac{1}{2} \left[ e^{\pi} (0) - e^{\pi/2} (1) \right] = \frac{1}{2} e^{\pi/2}$