To rewrite \( \sqrt{63} \) as the product of an integer and another irrational square root, we first identify perfect squares that can be factored from 63.
The prime factorization of 63 is:
\[ 63 = 9 \times 7 = 3^2 \times 7 \]
Since 9 is a perfect square, we can take its square root:
\[ \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} = 3\sqrt{7} \]
Thus, the irrational square root \( \sqrt{63} \) rewritten as the product of an integer and another irrational square root in simplest form is:
\[ 3\sqrt{7} \]