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

Options

A. `2sqrt(2)`
B. `2(sqrt(2) + 1)`
C. 0
D. `2(sqrt(2) -1)`

Answer

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

### Step 1: Rewrite the integrand
We start by rewriting the expression under the square root. We know that:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the integrand:
\[ 1 - \sin 2x = 1 - 2 \sin x \cos x \]

### Step 2: Use the identity for \( \sin^2 x + \cos^2 x \)
We can use the identity \( \sin^2 x + \cos^2 x = 1 \) to express \( 1 - 2 \sin x \cos x \):
\[ 1 - 2 \sin x \cos x = (\sin^2 x + \cos^2 x) - 2 \sin x \cos x = (\sin x - \cos x)^2 \]
So, we have:
\[ \sqrt{1 - \sin 2x} = \sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x| \]

### Step 3: Determine the absolute value
On the interval \( [0, \frac{\pi}{2}] \), \( \sin x \) is less than or equal to \( \cos x \) for \( x \in [0, \frac{\pi}{4}] \) and greater than or equal to \( \cos x \) for \( x \in [\frac{\pi}{4}, \frac{\pi}{2}] \). Thus:
\[ |\sin x - \cos x| = \begin{cases} \cos x - \sin x & \text{for } x \in [0, \frac{\pi}{4}] \\ \sin x - \cos x & \text{for } x \in [\frac{\pi}{4}, \frac{\pi}{2}] \end{cases} \]

### Step 4: Split the integral
We can split the integral into two parts:
\[ I = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx \]

### Step 5: Evaluate the first integral
Calculating the first integral:
\[ \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_{0}^{\frac{\pi}{4}} = \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) - \left( \sin 0 + \cos 0 \right) \]
\[ = \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right) - (0 + 1) = \sqrt{2} - 1 \]

### Step 6: Evaluate the second integral
Calculating the second integral:
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx = \left[ -\cos x - \sin x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \left( -\cos \frac{\pi}{2} - \sin \frac{\pi}{2} \right) - \left( -\cos \frac{\pi}{4} - \sin \frac{\pi}{4} \right) \]
\[ = (0 - 1) - \left( -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \right) = -1 + \sqrt{2} \]

### Step 7: Combine the results
Now, we combine both parts:
\[ I = (\sqrt{2} - 1) + (-1 + \sqrt{2}) = 2\sqrt{2} - 2 \]

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