`int_(0)^(pi//4) cos^(2) x dx`

Answer

To evaluate the integral \( I = \int_{0}^{\frac{\pi}{4}} \cos^2 x \, dx \), we will use a trigonometric identity and then perform the integration step by step.

### Step 1: Use the trigonometric identity
We know that:
\[ \cos 2x = 2 \cos^2 x - 1 \]
From this, we can express \(\cos^2 x\) as:
\[ \cos^2 x = \frac{1 + \cos 2x}{2} \]

### Step 2: Substitute in the integral
Now we substitute this expression into the integral:
\[ I = \int_{0}^{\frac{\pi}{4}} \cos^2 x \, dx = \int_{0}^{\frac{\pi}{4}} \frac{1 + \cos 2x}{2} \, dx \]
This simplifies to:
\[ I = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} (1 + \cos 2x) \, dx \]

### Step 3: Split the integral
We can split the integral into two parts:
\[ I = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{4}} 1 \, dx + \int_{0}^{\frac{\pi}{4}} \cos 2x \, dx \right) \]

### Step 4: Evaluate the first integral
The first integral is straightforward:
\[ \int_{0}^{\frac{\pi}{4}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Step 5: Evaluate the second integral
For the second integral, we have:
\[ \int_{0}^{\frac{\pi}{4}} \cos 2x \, dx \]
The integral of \(\cos 2x\) is:
\[ \int \cos 2x \, dx = \frac{\sin 2x}{2} \]
Thus,
\[ \int_{0}^{\frac{\pi}{4}} \cos 2x \, dx = \left[ \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{4}} = \frac{\sin(\frac{\pi}{2})}{2} - \frac{\sin(0)}{2} = \frac{1}{2} - 0 = \frac{1}{2} \]

### Step 6: Combine the results
Now, substituting back into our expression for \(I\):
\[ I = \frac{1}{2} \left( \frac{\pi}{4} + \frac{1}{2} \right) = \frac{1}{2} \left( \frac{\pi}{4} + \frac{2}{4} \right) = \frac{1}{2} \cdot \frac{\pi + 2}{4} = \frac{\pi + 2}{8} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{\pi + 2}{8} \]