Differentiate $2x \cos 3x$ with respect to $x$. Hence, find $\int x \sin 3x \text{ d}x$.

1. Apply the product rule to differentiate the expression

Let $y = 2x \cos 3x$.
Using the product rule $\frac{\text{d}}{\text{d}x}(uv) = u \frac{\text{d}v}{\text{d}x} + v \frac{\text{d}u}{\text{d}x}$ with $u = 2x$ and $v = \cos 3x$:
$\frac{\text{d}y}{\text{d}x} = (2x)\frac{\text{d}}{\text{d}x}(\cos 3x) + (\cos 3x)\frac{\text{d}}{\text{d}x}(2x)$

2. Evaluate the derivatives and simplify

$\frac{\text{d}}{\text{d}x}(\cos 3x) = -3 \sin 3x$ and $\frac{\text{d}}{\text{d}x}(2x) = 2$.
$\frac{\text{d}y}{\text{d}x} = (2x)(-3 \sin 3x) + (\cos 3x)(2)$
$= 2 \cos 3x - 6x \sin 3x$

1. State the derivative from part (a)

From (a), $\frac{\text{d}}{\text{d}x}(2x \cos 3x) = 2 \cos 3x - 6x \sin 3x$.

2. Apply integration as the reverse of differentiation

Integrating both sides with respect to $x$:
$\int (2 \cos 3x - 6x \sin 3x) \text{ d}x = 2x \cos 3x$
$\int 2 \cos 3x \text{ d}x - 6 \int x \sin 3x \text{ d}x = 2x \cos 3x$

3. Rearrange and integrate the remaining term to find the required integral

$6 \int x \sin 3x \text{ d}x = \int 2 \cos 3x \text{ d}x - 2x \cos 3x$
$6 \int x \sin 3x \text{ d}x = \frac{2}{3} \sin 3x - 2x \cos 3x + c'$
$\int x \sin 3x \text{ d}x = \frac{1}{9} \sin 3x - \frac{1}{3} x \cos 3x + c$