`int e^x cos 2x dx`

Answer

To solve the integral \( \int e^x \cos(2x) \, dx \), we will use the method of integration by parts. Let's go through the steps systematically.

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

### Step 2: Apply Integration by Parts
Now we apply the integration by parts formula:
\[ \int e^x \cos(2x) \, dx = e^x \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \cdot e^x \, dx \]
This simplifies to:
\[ \int e^x \cos(2x) \, dx = \frac{1}{2} e^x \sin(2x) - \frac{1}{2} \int e^x \sin(2x) \, dx \]

### Step 3: Set Up a New Integral
Let’s denote the original integral as \( I \):
\[ I = \int e^x \cos(2x) \, dx \]
Thus, we have:
\[ I = \frac{1}{2} e^x \sin(2x) - \frac{1}{2} \int e^x \sin(2x) \, dx \]

### Step 4: Apply Integration by Parts Again
Now we need to solve the integral \( \int e^x \sin(2x) \, dx \) using integration by parts again. Set:
- \( u = e^x \) (thus \( du = e^x \, dx \))
- \( dv = \sin(2x) \, dx \) (thus \( v = -\frac{1}{2} \cos(2x) \))

Now applying integration by parts:
\[ \int e^x \sin(2x) \, dx = e^x \left(-\frac{1}{2} \cos(2x)\right) - \int -\frac{1}{2} \cos(2x) \cdot e^x \, dx \]
This simplifies to:
\[ \int e^x \sin(2x) \, dx = -\frac{1}{2} e^x \cos(2x) + \frac{1}{2} \int e^x \cos(2x) \, dx \]

### Step 5: Substitute Back
Now we substitute this back into our expression for \( I \):
\[ I = \frac{1}{2} e^x \sin(2x) - \frac{1}{2} \left(-\frac{1}{2} e^x \cos(2x) + \frac{1}{2} I\right) \]
This leads to:
\[ I = \frac{1}{2} e^x \sin(2x) + \frac{1}{4} e^x \cos(2x) - \frac{1}{4} I \]

### Step 6: Solve for \( I \)
Now, we can combine like terms:
\[ I + \frac{1}{4} I = \frac{1}{2} e^x \sin(2x) + \frac{1}{4} e^x \cos(2x) \]
This simplifies to:
\[ \frac{5}{4} I = \frac{1}{2} e^x \sin(2x) + \frac{1}{4} e^x \cos(2x) \]
Thus, we can solve for \( I \):
\[ I = \frac{4}{5} \left(\frac{1}{2} e^x \sin(2x) + \frac{1}{4} e^x \cos(2x)\right) \]
\[ I = \frac{4}{10} e^x \sin(2x) + \frac{1}{5} e^x \cos(2x) \]
\[ I = \frac{2}{5} e^x \sin(2x) + \frac{1}{5} e^x \cos(2x) \]

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