Find $\frac{\text{d}}{\text{d}x}(x \cos 2x)$. Find $\frac{\text{d}}{\text{d}x}(x^2 \sin 2x)$. Using the results to parts (i) and (ii), find $\int x^2 \cos 2x \text{ d}x$.

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

Let $u = x$ and $v = \cos 2x$.
$\frac{\text{d}}{\text{d}x}(uv) = u \frac{\text{d}v}{\text{d}x} + v \frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}}{\text{d}x}(x \cos 2x) = x \frac{\text{d}}{\text{d}x}(\cos 2x) + (\cos 2x) \frac{\text{d}}{\text{d}x}(x)$

2. Differentiate the components and simplify

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

1. Apply the product rule to differentiate the expression

Let $u = x^2$ and $v = \sin 2x$.
$\frac{\text{d}}{\text{d}x}(uv) = u \frac{\text{d}v}{\text{d}x} + v \frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}}{\text{d}x}(x^2 \sin 2x) = x^2 \frac{\text{d}}{\text{d}x}(\sin 2x) + \sin 2x \frac{\text{d}}{\text{d}x}(x^2)$

2. Differentiate the individual terms and simplify

$\frac{\text{d}}{\text{d}x}(\sin 2x) = 2\cos 2x$
$\frac{\text{d}}{\text{d}x}(x^2) = 2x$
$\frac{\text{d}}{\text{d}x}(x^2 \sin 2x) = x^2(2\cos 2x) + (\sin 2x)(2x)$
$= 2x^2 \cos 2x + 2x \sin 2x$

1. Use the result from part (ii) to express the integral of $x^2 \cos 2x$ in terms of another integral.

From part (ii), $\frac{\text{d}}{\text{d}x}(x^2 \sin 2x) = 2x \sin 2x + 2x^2 \cos 2x$.
Rearranging gives:
$2x^2 \cos 2x = \frac{\text{d}}{\text{d}x}(x^2 \sin 2x) - 2x \sin 2x$
Integrating both sides with respect to $x$:
$\int 2x^2 \cos 2x \text{ d}x = x^2 \sin 2x - \int 2x \sin 2x \text{ d}x$

2. Use the result from part (i) to evaluate the remaining integral.

From part (i), $\frac{\text{d}}{\text{d}x}(x \cos 2x) = \cos 2x - 2x \sin 2x$.
Rearranging gives:
$2x \sin 2x = \cos 2x - \frac{\text{d}}{\text{d}x}(x \cos 2x)$
Integrating both sides with respect to $x$:
$\int 2x \sin 2x \text{ d}x = \int \cos 2x \text{ d}x - x \cos 2x$
$\int 2x \sin 2x \text{ d}x = \frac{1}{2} \sin 2x - x \cos 2x$

3. Substitute the evaluated integral back into the equation from step 1.

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

4. Divide by 2 to find the final integral and add the constant of integration.

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