Evaluate `int e^(x) cos^(2) x dx`

Answer

To evaluate the integral \( \int e^x \cos^2 x \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
Let \( I = \int e^x \cos^2 x \, dx \).

Using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \), we can rewrite the integral as:
\[ I = \int e^x \left( \frac{1 + \cos 2x}{2} \right) \, dx = \frac{1}{2} \int e^x \, dx + \frac{1}{2} \int e^x \cos 2x \, dx \]

### Step 2: Evaluate the first integral
The first integral is straightforward:
\[ \frac{1}{2} \int e^x \, dx = \frac{1}{2} e^x + C_1 \]

### Step 3: Set up the second integral
Let \( I_1 = \int e^x \cos 2x \, dx \). We will need to use integration by parts for this integral.

### Step 4: Apply integration by parts
Using integration by parts, let:
- \( u = \cos 2x \) (thus \( du = -2 \sin 2x \, dx \))
- \( dv = e^x \, dx \) (thus \( v = e^x \))

Applying integration by parts:
\[ I_1 = e^x \cos 2x - \int e^x (-2 \sin 2x) \, dx \]
\[ I_1 = e^x \cos 2x + 2 \int e^x \sin 2x \, dx \]

### Step 5: Set up the new integral
Let \( I_2 = \int e^x \sin 2x \, dx \). We will also apply integration by parts here:
- \( u = \sin 2x \) (thus \( du = 2 \cos 2x \, dx \))
- \( dv = e^x \, dx \) (thus \( v = e^x \))

Applying integration by parts:
\[ I_2 = e^x \sin 2x - \int e^x (2 \cos 2x) \, dx \]
\[ I_2 = e^x \sin 2x - 2 I_1 \]

### Step 6: Substitute back
Now we have two equations:
1. \( I_1 = e^x \cos 2x + 2 I_2 \)
2. \( I_2 = e^x \sin 2x - 2 I_1 \)

Substituting \( I_2 \) into the equation for \( I_1 \):
\[ I_1 = e^x \cos 2x + 2 \left( e^x \sin 2x - 2 I_1 \right) \]
\[ I_1 = e^x \cos 2x + 2 e^x \sin 2x - 4 I_1 \]
\[ 5 I_1 = e^x \cos 2x + 2 e^x \sin 2x \]
\[ I_1 = \frac{1}{5} e^x \cos 2x + \frac{2}{5} e^x \sin 2x \]

### Step 7: Substitute \( I_1 \) back into \( I \)
Now substitute \( I_1 \) back into the expression for \( I \):
\[ I = \frac{1}{2} e^x + \frac{1}{2} I_1 \]
\[ I = \frac{1}{2} e^x + \frac{1}{2} \left( \frac{1}{5} e^x \cos 2x + \frac{2}{5} e^x \sin 2x \right) \]
\[ I = \frac{1}{2} e^x + \frac{1}{10} e^x \cos 2x + \frac{1}{5} e^x \sin 2x \]

### Step 8: Combine and simplify
Combining terms:
\[ I = \frac{5}{10} e^x + \frac{1}{10} e^x \cos 2x + \frac{2}{10} e^x \sin 2x \]
\[ I = \frac{1}{10} e^x (5 + \cos 2x + 2 \sin 2x) + C \]

### Final Answer
Thus, the final result is:
\[ \int e^x \cos^2 x \, dx = \frac{1}{10} e^x (5 + \cos 2x + 2 \sin 2x) + C \]