5. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with its left focus at $F$, and $P\left(x_{0}, y_{0}\right)$ is a point on the ellipse, and $x_{0}>0$. A tangent line is drawn from point $P$ to the circle $x^{2}+y^{2}=b^{2}$, intersecting the ellipse at a second point $Q$. Let $I$ be the incenter of $\triangle P F Q$, and $\angle P F Q = 2 \alpha$. Then $|F I| \cos \alpha=$ $\qquad$
5. $a$.

Let the eccentricity of the ellipse be $e, P Q$ is tangent to the circle at point $M, Q\left(x_{Q}, y_{Q}\right)$, as shown in Figure 4.
$$
\begin{array}{l}
\text { Then }|P M|=\sqrt{|P O|^{2}-|O M|^{2}} \\
=\sqrt{x_{0}^{2}+y_{0}^{2}-b^{2}} \\
=\sqrt{x_{0}^{2}+b^{2}\left(1-\frac{x_{0}^{2}}{a^{2}}\right)-b^{2}} \\
=e x_{0} .
\end{array}
$$

Similarly, $|Q M|=e x_{Q}$.
By the second definition of the ellipse,
$$
|P F|=a+e x_{0},|Q F|=a+e x_{Q} .
$$

Draw $I N \perp P F$ at point $N$.
Thus, $|I F| \cos \alpha=|F N|$
$$
=\frac{1}{2}(|P F|+|Q F|-|P Q|)=a \text {. }
$$