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

Answer

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

### Step 1: Use the identity for \( \cos^2 x \)
We can use the trigonometric identity:
\[ \cos^2 x = \frac{1 + \cos 2x}{2} \]
Thus, we rewrite the integral:
\[ \int e^x \cos^2 x \, dx = \int e^x \left(\frac{1 + \cos 2x}{2}\right) \, dx = \frac{1}{2} \int e^x \, dx + \frac{1}{2} \int e^x \cos 2x \, dx \]

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

### Step 3: Solve the second integral using integration by parts
Let \( I = \int e^x \cos 2x \, dx \). We will use integration by parts:
- Let \( u = \cos 2x \) and \( dv = e^x \, dx \).
- Then, \( du = -2 \sin 2x \, dx \) and \( v = e^x \).

Using integration by parts:
\[ I = uv - \int v \, du = e^x \cos 2x - \int e^x (-2 \sin 2x) \, dx \]
This simplifies to:
\[ I = e^x \cos 2x + 2 \int e^x \sin 2x \, dx \]

### Step 4: Solve the integral \( \int e^x \sin 2x \, dx \) using integration by parts again
Let \( J = \int e^x \sin 2x \, dx \):
- Let \( u = \sin 2x \) and \( dv = e^x \, dx \).
- Then, \( du = 2 \cos 2x \, dx \) and \( v = e^x \).

Using integration by parts:
\[ J = uv - \int v \, du = e^x \sin 2x - \int e^x (2 \cos 2x) \, dx \]
This simplifies to:
\[ J = e^x \sin 2x - 2I \]

### Step 5: Substitute \( J \) back into the equation for \( I \)
Now we have:
\[ I = e^x \cos 2x + 2 \left( e^x \sin 2x - 2I \right) \]
This leads to:
\[ I = e^x \cos 2x + 2e^x \sin 2x - 4I \]
Combining like terms gives:
\[ 5I = e^x \cos 2x + 2e^x \sin 2x \]
Thus, we find:
\[ I = \frac{1}{5} e^x \cos 2x + \frac{2}{5} e^x \sin 2x \]

### Step 6: Combine results
Now substituting \( I \) back into our original integral:
\[ \int e^x \cos^2 x \, dx = \frac{1}{2} e^x + \frac{1}{2} \left( \frac{1}{5} e^x \cos 2x + \frac{2}{5} e^x \sin 2x \right) \]
This simplifies to:
\[ \int e^x \cos^2 x \, dx = \frac{1}{2} e^x + \frac{1}{10} e^x \cos 2x + \frac{1}{5} e^x \sin 2x + C \]

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