Differentiate $x \sin x + \cos x$ with respect to $x$. Show that $\frac{\text{d}}{\text{d}x} \left( \frac{1}{3} x \sin^3 x \right) = \frac{1}{3} \sin^3 x - x \cos^3 x + x \cos x$. Using the results found in part (a) and part (b), find $\int (\sin^3 x - 3x \cos^3 x) \text{d}x$.

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

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

2. Differentiate the entire expression and simplify

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

1. Apply the product rule to differentiate the expression

$\frac{\text{d}}{\text{d}x} \left( \frac{1}{3} x \sin^3 x \right) = \frac{1}{3} \left[ \sin^3 x \cdot \frac{\text{d}}{\text{d}x}(x) + x \cdot \frac{\text{d}}{\text{d}x}(\sin^3 x) \right]$

2. Differentiate the terms using the chain rule

$= \frac{1}{3} \left[ \sin^3 x (1) + x (3 \sin^2 x \cos x) \right]$
$= \frac{1}{3} \sin^3 x + x \sin^2 x \cos x$

3. Use trigonometric identity to express the result in the required form

Substitute $\sin^2 x = 1 - \cos^2 x$ into the second term:
$= \frac{1}{3} \sin^3 x + x (1 - \cos^2 x) \cos x$
$= \frac{1}{3} \sin^3 x + x (\cos x - \cos^3 x)$
$= \frac{1}{3} \sin^3 x - x \cos^3 x + x \cos x$ (Shown)

1. Find the derivative from part (a)

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

2. Rearrange the result from part (b) to match the integrand

From (b), multiplying by 3:
$\frac{\text{d}}{\text{d}x} (x \sin^3 x) = \sin^3 x - 3x \cos^3 x + 3x \cos x$
$\sin^3 x - 3x \cos^3 x = \frac{\text{d}}{\text{d}x} (x \sin^3 x) - 3x \cos x$

3. Integrate both sides with respect to $x$

$\int (\sin^3 x - 3x \cos^3 x) \text{d}x = \int \left( \frac{\text{d}}{\text{d}x} (x \sin^3 x) - 3x \cos x \right) \text{d}x$
$= x \sin^3 x - 3 \int x \cos x \text{d}x$

4. Substitute the integral of $x \cos x$ using the result from part (a)

Since $\frac{\text{d}}{\text{d}x} (x \sin x + \cos x) = x \cos x$, we have $\int x \cos x \text{d}x = x \sin x + \cos x$.
Therefore,
$\int (\sin^3 x - 3x \cos^3 x) \text{d}x = x \sin^3 x - 3(x \sin x + \cos x) + c$
$= x \sin^3 x - 3x \sin x - 3\cos x + c$