To solve the integral $\int_0^\pi e^{\cos(t)} \sin(2t) \, dt$, we can approach it using integration by parts or substitutions. In this case, it's simpler to use substitution and some known properties of integrals. Let’s break it down step by step.

Step 1: Look for a potential substitution

We want to simplify the integral $\int_0^\pi e^{\cos(t)} \sin(2t) \, dt$. We notice that the expression contains $e^{\cos(t)}$ and $\sin(2t)$. Let’s start by using a trigonometric identity for $\sin(2t)$:

$\sin(2t) = 2\sin(t)\cos(t)$

Thus, the integral becomes:

$I = \int_0^\pi e^{\cos(t)} \cdot 2 \sin(t) \cos(t) \, dt$

Step 2: Use substitution

Let’s substitute $u = \cos(t)$, which simplifies the expression considerably. This substitution works because $du = -\sin(t) \, dt$, and we can also change the limits of integration accordingly:

• When $t = 0$, $u = \cos(0) = 1$
• When $t = \pi$, $u = \cos(\pi) = -1$

Now, rewrite the integral in terms of $u$:

$I = 2 \int_1^{-1} e^u \cdot (-du)$

The negative sign from $du = -\sin(t) \, dt$ allows us to swap the limits of integration:

$I = 2 \int_{-1}^1 e^u \, du$

Step 3: Solve the simplified integral

The integral of $e^u$ is straightforward:

$\int e^u \, du = e^u$

Now, apply the limits of integration:

$I = 2 \left[ e^u \right]_{-1}^1 = 2 \left( e^1 - e^{-1} \right)$

Step 4: Final answer

Thus, the value of the integral is:

$I = 2 \left( e - \frac{1}{e} \right)$

This can be further simplified if needed, but the final answer is:

$\boxed{2 \left( e - \frac{1}{e} \right)}$

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Would you like further details or clarification on any step? Here are some related questions:

1. How can you recognize when to use substitution in integrals?
2. Why did we use $u = \cos(t)$ as the substitution here?
3. What are some common integrals involving $e^x$ that are useful to know?
4. How does swapping the limits of integration affect the sign of the integral?
5. What are other methods for solving integrals involving trigonometric functions?

Tip: Substitution is a powerful technique when the integrand contains a function and its derivative, making the integral easier to solve.