`int e^(" sin x "). " sin 2x dx "`

Answer

To solve the integral \( \int e^{\sin x} \sin 2x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral using the double angle identity for sine:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the integral as:
\[ \int e^{\sin x} \sin 2x \, dx = \int e^{\sin x} (2 \sin x \cos x) \, dx \]
This simplifies to:
\[ 2 \int e^{\sin x} \sin x \cos x \, dx \]

### Step 2: Substitution
Next, we will use substitution. Let:
\[ t = \sin x \]
Then, differentiating both sides gives:
\[ dt = \cos x \, dx \quad \Rightarrow \quad \cos x \, dx = dt \]
Now, substituting \( t \) into the integral, we have:
\[ 2 \int e^{t} t \, dt \]

### Step 3: Integration by Parts
We will now apply integration by parts to the integral \( \int e^{t} t \, dt \). Let:
- \( u = t \) (first function)
- \( dv = e^{t} dt \) (second function)

Then, we differentiate and integrate:
- \( du = dt \)
- \( v = e^{t} \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int e^{t} t \, dt = t e^{t} - \int e^{t} \, dt \]
Now, we know that:
\[ \int e^{t} \, dt = e^{t} \]
So, substituting back gives:
\[ \int e^{t} t \, dt = t e^{t} - e^{t} + C \]

### Step 4: Substitute Back
Now, we substitute back \( t = \sin x \):
\[ \int e^{\sin x} \sin 2x \, dx = 2 \left( \sin x e^{\sin x} - e^{\sin x} \right) + C \]
This simplifies to:
\[ 2 e^{\sin x} \left( \sin x - 1 \right) + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ \int e^{\sin x} \sin 2x \, dx = 2 e^{\sin x} (\sin x - 1) + C \]