Differentiate $2x \cos 2x$ with respect to $x$. Hence, evaluate $\int_{0}^{\frac{\pi}{2}} (1 - 4x \sin 2x) \, dx$.

1. Apply the product rule to differentiate the expression

Let $y = 2x \cos 2x$.
Using the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$ with $u = 2x$ and $v = \cos 2x$:
$\frac{dy}{dx} = (2x)\frac{d}{dx}(\cos 2x) + (\cos 2x)\frac{d}{dx}(2x)$

2. Differentiate the components and simplify

$\frac{dy}{dx} = (2x)(-2 \sin 2x) + (\cos 2x)(2)$
$= -4x \sin 2x + 2 \cos 2x$
$= 2 \cos 2x - 4x \sin 2x$

1. Find the derivative of $2x \cos 2x$ from part (a)

$\frac{d}{dx}(2x \cos 2x) = 2(\cos 2x) + 2x(-2 \sin 2x)$
$= 2 \cos 2x - 4x \sin 2x$

2. Rearrange the derivative to express $-4x \sin 2x$ as the subject

$-4x \sin 2x = \frac{d}{dx}(2x \cos 2x) - 2 \cos 2x$

3. Substitute the expression into the integral and integrate

$\int_{0}^{\frac{\pi}{2}} (1 - 4x \sin 2x) \, dx = \int_{0}^{\frac{\pi}{2}} \left(1 + \frac{d}{dx}(2x \cos 2x) - 2 \cos 2x\right) \, dx$
$= \left[ x + 2x \cos 2x - \sin 2x \right]_{0}^{\frac{\pi}{2}}$

4. Evaluate the definite integral using the limits

$= \left( \frac{\pi}{2} + 2\left(\frac{\pi}{2}\right) \cos \pi - \sin \pi \right) - (0 + 0 - 0)$
$= \frac{\pi}{2} + \pi(-1) - 0$
$= -\frac{\pi}{2}$