To rewrite \( \sqrt{63} \) as the product of an integer and another irrational square root, we start by factoring 63 into its prime components:
\[ 63 = 9 \times 7 = 3^2 \times 7 \]
Now, we can take the square root of both factors:
\[ \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} \]
Since \( \sqrt{9} = 3 \), we have:
\[ \sqrt{63} = 3 \times \sqrt{7} \]
Thus, the square root \( \sqrt{63} \) can be rewritten in simpler form as:
\[ \sqrt{63} = 3\sqrt{7} \]
In conclusion, the answer is:
\[ \sqrt{63} = 3\sqrt{7} \]