2. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
[\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{3+\sqrt{2}}}}}]
$$

is equal to
(there are a total of 2009 square roots).
2. 45 .

Let $A=\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{3+\sqrt{2}}}}}$.
Then $A>\sqrt{2010+\sqrt{2009}}$
$$
\begin{array}{l}
>\sqrt{2010+44}>45, \\
A<\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{2009}}}} \\
<\sqrt{2010+46}<46 .
\end{array}
$$

Thus, $[A]=45$.