$$ \int _ { 0 } ^ { \pi } e ^ { \cos ( t ) } \sin ( 2 t ) d t $$

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To evaluate the integral

$$
\int_{0}^{\pi} e^{\cos(t)} \sin(2t) \, dt,
$$

we can use the method of integration by parts. Let's proceed step by step.

### Step 1: Integration by Parts

Let’s set:
- $$ u = e^{\cos(t)} $$ 
- $$ dv = \sin(2t) \, dt $$

Then, we compute:
- $$ du = -e^{\cos(t)} \sin(t) \, dt $$
- $$ v = -\frac{\cos(2t)}{2} $$

Applying integration by parts:

$$
\int u \, dv = uv - \int v \, du
$$

Substituting the values:

$$
\int_{0}^{\pi} e^{\cos(t)} \sin(2t) \, dt = \left[ -\frac{e^{\cos(t)} \cos(2t)}{2} \right]_0^{\pi} - \int_{0}^{\pi} \left(-\frac{\cos(2t)}{2}\right) \left(-e^{\cos(t)} \sin(t)\right) \, dt
$$

Simplifying the boundary terms:

$$
\left[ -\frac{e^{\cos(\pi)} \cos(2\pi)}{2} + \frac{e^{\cos(0)} \cos(0)}{2} \right] = \frac{e - e^{-1}}{2}
$$

Now, the remaining integral becomes:

$$
-\frac{1}{2} \int_{0}^{\pi} \cos(2t) e^{\cos(t)} \sin(t) \, dt
$$

### Step 2: Substitution

Let $$ u = \cos(t) $$, hence $$ du = -\sin(t) \, dt $$.

The limits change as follows:
- When $$ t = 0 $$, $$ u = 1 $$
- When $$ t = \pi $$, $$ u = -1 $$

Substituting, we get:

$$
\int_{0}^{\pi} \cos(2t) e^{\cos(t)} \sin(t) \, dt = \int_{-1}^{1} (2u^2 - 1) e^{u} \, du
$$

This integral can be evaluated as:

$$
\int_{-1}^{1} (2u^2 - 1) e^{u} \, du = e - \frac{9}{e}
$$

### Step 3: Combine the Results

Putting everything back together:

$$
\frac{e - e^{-1}}{2} - \frac{1}{2} \left( e - \frac{9}{e} \right) = \frac{4}{e}
$$

### Final Answer

$$
\boxed{\dfrac{4}{e}}
$$
The value of the integral is $$ \frac{4}{e} $$.

\( \int _ { 0 } ^ { \pi } e ^ { \cos ( t ) } \sin ( 2 t ) d t \)

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