The domain of \(\sin^{-1}(x)\) is \([-1, 1]\). Therefore, for \(\sin^{-1}\sqrt{x-1}\) to be defined, we must have:

\(-1 \le \sqrt{x-1} \le 1\)

Since the square root is always non-negative, we can simplify the inequality to:

\(0 \le \sqrt{x-1} \le 1\)

Squaring all parts of the inequality, we get:

\(0 \le x-1 \le 1\)

Adding 1 to all parts of the inequality, we get:

\(1 \le x \le 2\)

Therefore, the domain of \(\sin^{-1}\sqrt{x-1}\) is \([1, 2]\).