To solve the integral
\[
I = \int_{0}^{\pi} \frac{e^{|\cos x|} \sin x}{1 + e^{\cot x}} \, dx,
\]
we can use the property of definite integrals which states that
\[
\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx.
\]
In our case, \( a = 0 \) and \( b = \pi \), so we have:
\[
I = \int_{0}^{\pi} f(\pi - x) \, dx.
\]
First, we need to compute \( f(\pi - x) \):
1. **Calculate \( f(\pi - x) \)**:
\[
f(\pi - x) = \frac{e^{|\cos(\pi - x)|} \sin(\pi - x)}{1 + e^{\cot(\pi - x)}}
\]
Since \( \cos(\pi - x) = -\cos x \), we have \( |\cos(\pi - x)| = |\cos x| \) and \( \sin(\pi - x) = \sin x \). Also, \( \cot(\pi - x) = -\cot x \).
Therefore,
\[
f(\pi - x) = \frac{e^{|\cos x|} \sin x}{1 + e^{-\cot x}}.
\]
2. **Rewrite \( f(\pi - x) \)**:
\[
f(\pi - x) = \frac{e^{|\cos x|} \sin x}{1 + \frac{1}{e^{\cot x}}} = \frac{e^{|\cos x|} \sin x \cdot e^{\cot x}}{e^{\cot x} + 1}.
\]
3. **Add \( I \) and \( f(\pi - x) \)**:
Now, we can add \( I \) and \( f(\pi - x) \):
\[
2I = \int_{0}^{\pi} \left( f(x) + f(\pi - x) \right) \, dx.
\]
This gives:
\[
2I = \int_{0}^{\pi} \left( \frac{e^{|\cos x|} \sin x}{1 + e^{\cot x}} + \frac{e^{|\cos x|} \sin x \cdot e^{\cot x}}{e^{\cot x} + 1} \right) \, dx.
\]
4. **Simplify the expression**:
The denominators are the same, so we can combine the fractions:
\[
2I = \int_{0}^{\pi} \frac{e^{|\cos x|} \sin x (1 + e^{\cot x})}{1 + e^{\cot x}} \, dx.
\]
This simplifies to:
\[
2I = \int_{0}^{\pi} e^{|\cos x|} \sin x \, dx.
\]
5. **Evaluate the integral**:
Now we can evaluate:
\[
I = \frac{1}{2} \int_{0}^{\pi} e^{|\cos x|} \sin x \, dx.
\]
6. **Change of variable**:
Let \( t = \cos x \), then \( dt = -\sin x \, dx \). The limits change from \( x = 0 \) (where \( t = 1 \)) to \( x = \pi \) (where \( t = -1 \)):
\[
I = \frac{1}{2} \int_{1}^{-1} e^{|t|} (-dt) = \frac{1}{2} \int_{-1}^{1} e^{|t|} \, dt.
\]
7. **Split the integral**:
We can split this integral into two parts:
\[
I = \frac{1}{2} \left( \int_{-1}^{0} e^{-t} \, dt + \int_{0}^{1} e^{t} \, dt \right).
\]
8. **Evaluate the integrals**:
- For \( \int_{-1}^{0} e^{-t} \, dt \):
\[
= \left[ -e^{-t} \right]_{-1}^{0} = -e^{0} + e^{1} = 1 - e.
\]
- For \( \int_{0}^{1} e^{t} \, dt \):
\[
= \left[ e^{t} \right]_{0}^{1} = e - 1.
\]
9. **Combine results**:
Thus,
\[
I = \frac{1}{2} \left( (1 - e) + (e - 1) \right) = \frac{1}{2} \cdot 2(e - 1) = e - 1.
\]
Therefore, the value of the integral is
\[
\boxed{e - 1}.
\]
