Step 1: Fix $x$ and substitute inside the inner integral.
For a fixed $x$, let $u = xy$, so $du = x\,dy$.
When $y = 0$, $u = 0$; when $y = 1$, $u = x$.
$\displaystyle\int_0^1 x\cos(xy)\,dy = \int_0^{x} \cos(u)\,du = \sin(x) - \sin(0) = \sin(x)$.
Step 2: Integrate this result over $x$.
$\displaystyle I = \int_0^1 \sin(x)\,dx$.
The antiderivative of $\sin(x)$ is $-\cos(x)$, so $I = [-\cos(x)]_0^1 = \cos(0) - \cos(1) = 1 - \cos(1)$.
Step 3: Rule out the distractors.
Adding instead of subtracting cosine, as in option B, gives a value bigger than 1, which cannot happen since $\sin(x) \le 1$ over the interval.
Options C and D wrongly bring in $\sin(1)$ where $\cos(1)$ belongs.
Final Answer:
The substitution method gives the same closed form, $1 - \cos(1)$, confirming option A.
\[ \boxed{I = 1 - \cos(1)} \]