5. Given the function $f(x)=\mathrm{e}^{x}(\sin x+\cos x)$, where $x \in\left[-\frac{2011 \pi}{2}, \frac{2013 \pi}{2}\right]$. A tangent line is drawn to the graph of the function $f(x)$ through the point $M\left(\frac{\pi-1}{2}, 0\right)$. Let the x-coordinates of the points of tangency form the sequence $\left\{x_{n}\right\}$. Then the sum $S$ of all terms in the sequence $\left\{x_{n}\right\}$ is $\qquad$
5. $1006 \pi$.

Let the coordinates of the tangent point be $\left(x_{0}, \mathrm{e}^{x_{0}}\left(\sin x_{0}+\cos x_{0}\right)\right)$.
Then the equation of the tangent line is
$$
y-\mathrm{e}^{x_{0}}\left(\sin x_{0}+\cos x_{0}\right)=2 \mathrm{e}^{x_{0}} \cos x_{0} \cdot\left(x-x_{0}\right) \text {. }
$$

Substituting the coordinates of point $M\left(\frac{\pi-1}{2}, 0\right)$ into the tangent line equation, we get
$$
\begin{array}{l}
-\mathrm{e}^{x_{0}}\left(\sin x_{0}+\cos x_{0}\right)=2 \mathrm{e}^{x_{0}} \cos x_{0} \cdot\left(\frac{\pi-1}{2}-x_{0}\right) \\
\Rightarrow \tan x_{0}+1=2\left(x_{0}-\frac{\pi-1}{2}\right) \\
\Rightarrow \tan x_{0}=2\left(x_{0}-\frac{\pi}{2}\right) . \\
\text { Let } y_{1}=\tan x, y_{2}=2\left(x-\frac{\pi}{2}\right) .
\end{array}
$$

Then the graphs of these two functions are both symmetric about the point $\left(\frac{\pi}{2}, 0\right)$, and the x-coordinates of their intersection points are also symmetric about $x=\frac{\pi}{2}$ in pairs. The roots of the equation
$$
\tan x=2\left(x-\frac{\pi}{2}\right)\left(x \in\left[-\frac{2011 \pi}{2}, \frac{2013 \pi}{2}\right]\right)
$$

are the x-coordinates of the tangent points of all the tangents drawn, and they form a sequence $\left\{x_{n}\right\}$ that is symmetric about $x=\frac{\pi}{2}$ in pairs. Within $\left[-\frac{2011 \pi}{2}, \frac{2013 \pi}{2}\right)$, there are 1006 pairs, and the sum of each pair is $\pi$. Therefore, the sum $S$ of all terms in the sequence $\left\{x_{n}\right\}$ is $S=1006 \pi$.