We are asked to evaluate the definite integral:

\[
I = \int_0^\pi e^{\sin(t)} \cos(t) \sin(t) \, dt
\]

To solve this, notice that we can rewrite the product of \(\cos(t) \sin(t)\) as:

\[
\cos(t) \sin(t) = \frac{1}{2} \sin(2t)
\]

Thus, the integral becomes:

\[
I = \int_0^\pi e^{\sin(t)} \cdot \frac{1}{2} \sin(2t) \, dt
\]

Now let's simplify this integral and examine potential substitutions:

1. Set \( u = \sin(t) \), which implies \( du = \cos(t) dt \).
2. The limits change accordingly: when \( t = 0 \), \( u = \sin(0) = 0 \), and when \( t = \pi \), \( u = \sin(\pi) = 0 \).
3. The integral becomes more manageable. However, due to the symmetry of sine over the interval, the result will be zero, since the integrand is an odd function (because of the \(\sin(2t)\)) over a symmetric interval from \(0\) to \(\pi\).

Thus, the value of the integral is:

\[
I = 0
\]

Would you like more details on any of the steps? 

Here are 5 questions to consider:
1. What are the properties of definite integrals that help simplify symmetric intervals?
2. How can trigonometric identities simplify integrals involving sine and cosine?
3. What other techniques could be applied to solve integrals involving exponentials and trigonometric functions?
4. Why does the function \(\sin(2t)\) being odd lead to the integral being zero?
5. How would the integral change if the limits were different?

**Tip:** Symmetry in integrals is a powerful tool—always check if the integrand is odd or even when dealing with symmetric limits.

Explore the solution to the definite integral ∫_0^π e^sin(t) cos(t) sin(t) dt using trigonometric identities and symmetry properties. Learn how the integral simplifies due to the odd nature of the function over a symmetric interval, resulting in a value of zero. Suitable for university-level calculus students.

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