`int_(0)^(pi//2)sqrt( 1- sin 2 x ) dx` is equal to

Options

A. `2 sqrt(2)`
B. `2 ( sqrt(2) +1)`
C. 2
D. `2 ( sqrt(2) -1)`

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \sin 2x} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We start with the expression inside the square root:
\[ 1 - \sin 2x \]
Using the double angle identity, we know that \( \sin 2x = 2 \sin x \cos x \). Thus, we can rewrite:
\[ 1 - \sin 2x = 1 - 2 \sin x \cos x \]

### Step 2: Use the identity for squares
We can express \( 1 - 2 \sin x \cos x \) as:
\[ 1 - 2 \sin x \cos x = (\sin x - \cos x)^2 \]
This follows from the identity \( a^2 + b^2 - 2ab = (a - b)^2 \). Here, \( a = \sin x \) and \( b = \cos x \).

### Step 3: Rewrite the integral
Now we can rewrite the integral:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \sin 2x} \, dx = \int_{0}^{\frac{\pi}{2}} \sqrt{(\sin x - \cos x)^2} \, dx \]
Since \( \sin x - \cos x \) is non-negative in the interval \( [0, \frac{\pi}{2}] \), we can simplify:
\[ \sqrt{(\sin x - \cos x)^2} = \sin x - \cos x \]
Thus, the integral becomes:
\[ \int_{0}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx \]

### Step 4: Integrate
Now we can integrate:
\[ \int (\sin x - \cos x) \, dx = -\cos x - \sin x + C \]
Evaluating from \( 0 \) to \( \frac{\pi}{2} \):
\[ \left[-\cos x - \sin x\right]_{0}^{\frac{\pi}{2}} = \left[-\cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)\right] - \left[-\cos(0) - \sin(0)\right] \]
Calculating the values:
\[ = \left[-0 - 1\right] - \left[-1 - 0\right] = -1 + 1 = 0 \]

### Final Result
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \sin 2x} \, dx = 0 \]