`int e^x sin 2x dx`

Answer

To solve the integral \( I = \int e^x \sin(2x) \, dx \), we will use the method of integration by parts. Let's break it down step by step.

### Step 1: Set up the integration by parts
We will use the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Here, we choose:
- \( u = e^x \) (which means \( du = e^x \, dx \))
- \( dv = \sin(2x) \, dx \) (which means \( v = -\frac{1}{2} \cos(2x) \))

### Step 2: Apply the integration by parts formula
Substituting into the integration by parts formula:
\[ I = e^x \left(-\frac{1}{2} \cos(2x)\right) - \int \left(-\frac{1}{2} \cos(2x)\right) e^x \, dx \]
This simplifies to:
\[ I = -\frac{1}{2} e^x \cos(2x) + \frac{1}{2} \int e^x \cos(2x) \, dx \]

### Step 3: Set up the new integral
Let’s denote the new integral as \( J = \int e^x \cos(2x) \, dx \). Now we have:
\[ I = -\frac{1}{2} e^x \cos(2x) + \frac{1}{2} J \]

### Step 4: Solve the new integral \( J \)
We will again use integration by parts for \( J \):
- Choose \( u = e^x \) (thus \( du = e^x \, dx \))
- Choose \( dv = \cos(2x) \, dx \) (thus \( v = \frac{1}{2} \sin(2x) \))

Applying the integration by parts formula:
\[ J = e^x \left(\frac{1}{2} \sin(2x)\right) - \int \left(\frac{1}{2} \sin(2x)\right) e^x \, dx \]
This simplifies to:
\[ J = \frac{1}{2} e^x \sin(2x) - \frac{1}{2} I \]

### Step 5: Substitute \( J \) back into the equation for \( I \)
Now we substitute \( J \) back into the equation for \( I \):
\[ I = -\frac{1}{2} e^x \cos(2x) + \frac{1}{2} \left(\frac{1}{2} e^x \sin(2x) - \frac{1}{2} I\right) \]
This simplifies to:
\[ I = -\frac{1}{2} e^x \cos(2x) + \frac{1}{4} e^x \sin(2x) - \frac{1}{4} I \]

### Step 6: Combine like terms
Now, we can combine the terms involving \( I \):
\[ I + \frac{1}{4} I = -\frac{1}{2} e^x \cos(2x) + \frac{1}{4} e^x \sin(2x) \]
This simplifies to:
\[ \frac{5}{4} I = -\frac{1}{2} e^x \cos(2x) + \frac{1}{4} e^x \sin(2x) \]

### Step 7: Solve for \( I \)
Now, we can solve for \( I \):
\[ I = \frac{4}{5} \left(-\frac{1}{2} e^x \cos(2x) + \frac{1}{4} e^x \sin(2x)\right) \]
This simplifies to:
\[ I = -\frac{2}{5} e^x \cos(2x) + \frac{1}{5} e^x \sin(2x) \]

### Step 8: Final answer
Thus, the integral is:
\[ I = \frac{1}{5} e^x \sin(2x) - \frac{2}{5} e^x \cos(2x) + C \]
where \( C \) is the constant of integration.