It is given that $y = 4x \sin 2x$.

1. Apply the product rule to differentiate $y = 4x \sin 2x$

Let $u = 4x$ and $v = \sin 2x$.
Then $\frac{\mathrm{d}u}{\mathrm{d}x} = 4$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = 2 \cos 2x$.
Using the product rule $\frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x}$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = (4x)(2 \cos 2x) + (\sin 2x)(4)$

2. Simplify the expression to the required form

$\frac{\mathrm{d}y}{\mathrm{d}x} = 8x \cos 2x + 4 \sin 2x$
(This is in the form $ax \cos 2x + b \sin 2x$ with $a = 8$ and $b = 4$)

1. Evaluate the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $x = \frac{\pi}{2}$ using the result from part (i).

From part (i), $\frac{\mathrm{d}y}{\mathrm{d}x} = 8x \cos 2x + 4 \sin 2x$.
When $x = \frac{\pi}{2}$,
$\frac{\mathrm{d}y}{\mathrm{d}x} = 8\left(\frac{\pi}{2}\right) \cos\left(2 \times \frac{\pi}{2}\right) + 4 \sin\left(2 \times \frac{\pi}{2}\right)$
$= 4\pi \cos \pi + 4 \sin \pi$
$= 4\pi(-1) + 0 = -4\pi$

2. Apply the chain rule to find the rate of change of $y$ with respect to time.

Given $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{2}{5}$.
Using the chain rule, $\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}$
$\frac{\mathrm{d}y}{\mathrm{d}t} = -4\pi \times \frac{2}{5} = -\frac{8\pi}{5}$

1. State the derivative from part (i) and express it as an integral equation

From part (i), $\frac{\mathrm{d}}{\mathrm{d}x}(4x \sin 2x) = 8x \cos 2x + 4 \sin 2x$.
Integrating both sides with respect to $x$:
$\int (8x \cos 2x + 4 \sin 2x) \mathrm{d}x = 4x \sin 2x + c$
$8 \int x \cos 2x \mathrm{d}x + \int 4 \sin 2x \mathrm{d}x = 4x \sin 2x + c$

2. Rearrange to make the required integral the subject and apply limits

$8 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} x \cos 2x \mathrm{d}x = \left[ 4x \sin 2x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 4 \sin 2x \mathrm{d}x$

3. Evaluate the integrated terms

$\left[ 4x \sin 2x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \left( 4\left(\frac{\pi}{2}\right) \sin \pi \right) - \left( 4\left(\frac{\pi}{4}\right) \sin \frac{\pi}{2} \right)$
$= 0 - \pi(1) = -\pi$

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

4. Solve for the final integral

$8 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} x \cos 2x \mathrm{d}x = -\pi - 2$
$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} x \cos 2x \mathrm{d}x = \frac{-\pi - 2}{8} = -\frac{\pi + 2}{8}$

1. Interpret the negative result of the definite integral from part (iii)

Since the evaluated definite integral $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} x \cos 2x \mathrm{d}x$ is negative, the area bounded by the curve and the $x$-axis in this interval lies below the $x$-axis.