Step 1: Tabular method
Differentiate $x^3$ repeatedly: $x^3, 3x^2, 6x, 6, 0$. Integrate $\cos x$ repeatedly: $\sin x, -\cos x, -\sin x, \cos x$.
Step 2: Alternate signs
Pair them with signs $+, -, +, -$: $+x^3\sin x$, $-3x^2(-\cos x)$, $+6x(-\sin x)$, $-6(\cos x)$.
Step 3: Sum
$x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + c$.
Step 4: Verify by differentiation
Differentiate term by term: $x^3\sin x \to 3x^2\sin x + x^3\cos x$; $3x^2\cos x \to 6x\cos x - 3x^2\sin x$; $-6x\sin x \to -6\sin x - 6x\cos x$; $-6\cos x \to 6\sin x$. Everything cancels except $x^3\cos x$. Options A, B and D differ in a sign or a trig function, so their derivatives leave unwanted terms.
Final Answer:
The result is option (C). This is option (C).
\[ \boxed{\text{(C) }x^3\sin x+3x^2\cos x-6x\sin x-6\cos x+c} \]