The value of the integral `int_(-pi/2)^(pi//2) sqrt((1+cos2x)/(2))dx` is

Options

A. -2
B. 2
C. 0
D. -3

Answer

To solve the integral \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{1 + \cos 2x}{2}} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
We can use the trigonometric identity for \( \cos 2x \):
\[ \cos 2x = 2\cos^2 x - 1 \]
Thus, we can rewrite \( 1 + \cos 2x \) as:
\[ 1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x \]
Now substituting this back into the integral:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{2\cos^2 x}{2}} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\cos^2 x} \, dx \]

### Step 2: Simplify the square root
Since \( \sqrt{\cos^2 x} = |\cos x| \), we need to consider the absolute value of \( \cos x \) over the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). In this interval, \( \cos x \) is non-negative:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx \]

### Step 3: Use the property of even functions
The function \( \cos x \) is an even function, which means:
\[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \]
Thus, we can write:
\[ I = 2 \int_{0}^{\frac{\pi}{2}} \cos x \, dx \]

### Step 4: Evaluate the integral
Now we evaluate the integral:
\[ \int \cos x \, dx = \sin x \]
So we have:
\[ I = 2 \left[ \sin x \right]_{0}^{\frac{\pi}{2}} = 2 \left( \sin\left(\frac{\pi}{2}\right) - \sin(0) \right) = 2 \left( 1 - 0 \right) = 2 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{2} \]