` int_(0)^(pi) sqrt((cos2x+1)/2)dx` is equal to

Options

A. 0
B. 2
C. -2
D. None of these

Explanation

To evaluate the integral

\[ I = \int_{0}^{\pi} \sqrt{\frac{1 + \cos 2x}{2}} \, dx, \]

we can follow these steps:

### Step 1: Simplify the integrand using trigonometric identities
We know that:

\[ 1 + \cos 2x = 2 \cos^2 x. \]

Thus, we can rewrite the integrand:

\[ \sqrt{\frac{1 + \cos 2x}{2}} = \sqrt{\frac{2 \cos^2 x}{2}} = \sqrt{\cos^2 x} = |\cos x|. \]

### Step 2: Rewrite the integral
Now, we can express the integral as:

\[ I = \int_{0}^{\pi} |\cos x| \, dx. \]

### Step 3: Analyze the absolute value
The function \(\cos x\) is non-negative in the interval \([0, \frac{\pi}{2}]\) and non-positive in the interval \([\frac{\pi}{2}, \pi]\). Therefore, we can split the integral into two parts:

\[ I = \int_{0}^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx. \]

### Step 4: Evaluate the first integral
The first integral is:

\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx. \]

The antiderivative of \(\cos x\) is \(\sin x\), so we have:

\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = \left[ \sin x \right]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1. \]

### Step 5: Evaluate the second integral
The second integral is:

\[ \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx. \]

Again, using the antiderivative of \(-\cos x\), we have:

\[ \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx = -\left[ \sin x \right]_{\frac{\pi}{2}}^{\pi} = -(\sin(\pi) - \sin\left(\frac{\pi}{2}\right)) = -\left(0 - 1\right) = 1. \]

### Step 6: Combine the results
Now, we can combine the results of both integrals:

\[ I = 1 + 1 = 2. \]

### Final Answer
Thus, the value of the integral is:

\[ \boxed{2}. \]