`int_(-pi//2)^(pi//2) sqrt((1-cos 2x)/( 2)) dx` is equal to

Options

A. 0
B. `(1)/(2)`
C. 1
D. 2

Answer

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

### Step 1: Simplify the integrand
We start with the expression inside the integral:
\[ \sqrt{\frac{1 - \cos 2x}{2}} \]
Using the trigonometric identity \( \cos 2x = 1 - 2\sin^2 x \), we can rewrite \( 1 - \cos 2x \):
\[ 1 - \cos 2x = 2\sin^2 x \]
Thus, we can substitute this back into our integral:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{2\sin^2 x}{2}} \, dx \]

### Step 2: Simplify further
The expression simplifies to:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\sin^2 x} \, dx \]
Since \( \sqrt{\sin^2 x} = |\sin x| \), we have:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |\sin x| \, dx \]

### Step 3: Evaluate the integral
The function \( |\sin x| \) is symmetric about the y-axis and is positive in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). Therefore, we can write:
\[ I = 2 \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

### Step 4: Compute the integral
Now we compute the integral \( \int_{0}^{\frac{\pi}{2}} \sin x \, dx \):
\[ \int \sin x \, dx = -\cos x \]
Evaluating from \( 0 \) to \( \frac{\pi}{2} \):
\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = [-\cos x]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = 0 + 1 = 1 \]

### Step 5: Final result
Thus, we have:
\[ I = 2 \cdot 1 = 2 \]

### Conclusion
The value of the integral \( I \) is:
\[ \boxed{2} \]