Differentiate $3x \sin 2x$ with respect to $x$. Hence, evaluate $\int_{0}^{\frac{\pi}{4}} x \cos 2x \text{ d}x$.

1. Apply the product rule to differentiate $3x \sin 2x$

Let $u = 3x$ and $v = \sin 2x$.
$\frac{du}{dx} = 3$
$\frac{dv}{dx} = 2 \cos 2x$
$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

2. Substitute and simplify the expression

$\frac{d}{dx}(3x \sin 2x) = (3x)(2 \cos 2x) + (\sin 2x)(3)$
$= 6x \cos 2x + 3 \sin 2x$

1. Find the derivative of $3x \sin 2x$ as required by part (i).

Let $y = 3x \sin 2x$.
Using the product rule:
$\frac{\text{d}y}{\text{d}x} = 3(\sin 2x) + 3x(2 \cos 2x)$
$\frac{\text{d}y}{\text{d}x} = 3 \sin 2x + 6x \cos 2x$

2. Use integration as the reverse of differentiation to express the integral of $x \cos 2x$.

Since $\frac{\text{d}}{\text{d}x}(3x \sin 2x) = 3 \sin 2x + 6x \cos 2x$, integrating both sides gives:
$\int (3 \sin 2x + 6x \cos 2x) \text{ d}x = 3x \sin 2x$
$\int 6x \cos 2x \text{ d}x = 3x \sin 2x - \int 3 \sin 2x \text{ d}x$
Dividing by 6:
$\int x \cos 2x \text{ d}x = \frac{1}{2}x \sin 2x - \frac{1}{2} \int \sin 2x \text{ d}x$

3. Integrate the remaining trigonometric term.

$\int x \cos 2x \text{ d}x = \frac{1}{2}x \sin 2x - \frac{1}{2} \left( -\frac{1}{2} \cos 2x \right)$
$\int x \cos 2x \text{ d}x = \frac{1}{2}x \sin 2x + \frac{1}{4} \cos 2x$

4. Evaluate the definite integral using the given limits.

$\int_{0}^{\frac{\pi}{4}} x \cos 2x \text{ d}x = \left[ \frac{1}{2}x \sin 2x + \frac{1}{4} \cos 2x \right]_{0}^{\frac{\pi}{4}}$
$= \left( \frac{1}{2}\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{2}\right) + \frac{1}{4} \cos\left(\frac{\pi}{2}\right) \right) - \left( 0 + \frac{1}{4} \cos(0) \right)$
$= \left( \frac{\pi}{8}(1) + 0 \right) - \left( \frac{1}{4}(1) \right)$
$= \frac{\pi}{8} - \frac{1}{4}$