Evaluate : `int_(0)^(pi) cos^(2)x.dx`

Answer

To evaluate the integral \( \int_{0}^{\pi} \cos^2 x \, dx \), we can follow these steps:

### Step 1: Use the identity for \( \cos^2 x \)
We know from trigonometric identities that:
\[ \cos^2 x = \frac{1 + \cos 2x}{2} \]
Thus, we can rewrite the integral as:
\[ \int_{0}^{\pi} \cos^2 x \, dx = \int_{0}^{\pi} \frac{1 + \cos 2x}{2} \, dx \]

### Step 2: Factor out the constant
We can factor out \( \frac{1}{2} \) from the integral:
\[ = \frac{1}{2} \int_{0}^{\pi} (1 + \cos 2x) \, dx \]

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

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

### Step 5: Evaluate the second integral
For the second integral, we use the fact that:
\[ \int \cos kx \, dx = \frac{1}{k} \sin kx + C \]
Thus,
\[ \int_{0}^{\pi} \cos 2x \, dx = \frac{1}{2} \sin 2x \bigg|_{0}^{\pi} = \frac{1}{2} (\sin 2\pi - \sin 0) = \frac{1}{2} (0 - 0) = 0 \]

### Step 6: Combine the results
Now, substituting back into our expression:
\[ = \frac{1}{2} \left( \pi + 0 \right) = \frac{\pi}{2} \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{\pi} \cos^2 x \, dx \) is:
\[ \frac{\pi}{2} \]

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