Choose the correct Answer of the Following Questions : `int_(pi/4)^(pi/2) sqrt(1-sin2x) dx` is equal to

Options

A. `sqrt2 + 1`
B. `2(sqrt2 - 1)`
C. `1- sqrt2`
D. 0

Answer

To solve the integral \( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 - \sin 2x} \, dx \), we will 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 for sine, we know:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite:
\[ 1 - \sin 2x = 1 - 2 \sin x \cos x \]

### Step 2: Rewrite \(1\)
We can express \(1\) as:
\[ 1 = \sin^2 x + \cos^2 x \]
So, we have:
\[ 1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x \]
This can be recognized as a perfect square:
\[ 1 - \sin 2x = (\cos x - \sin x)^2 \]

### Step 3: Substitute back into the integral
Now substituting back into the integral, we have:
\[ \sqrt{1 - \sin 2x} = \sqrt{(\cos x - \sin x)^2} = |\cos x - \sin x| \]
Since \(x\) is in the interval \([\frac{\pi}{4}, \frac{\pi}{2}]\), \(\cos x - \sin x\) is negative. Therefore:
\[ \sqrt{1 - \sin 2x} = \sin x - \cos x \]

### Step 4: Set up the integral
Now we can rewrite the integral:
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 - \sin 2x} \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx \]

### Step 5: Integrate
Now we can integrate:
\[ \int (\sin x - \cos x) \, dx = -\cos x - \sin x + C \]
Thus, we evaluate the definite integral:
\[ \left[-\cos x - \sin x\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \]

### Step 6: Evaluate at the limits
Now we calculate at the limits:
1. At \(x = \frac{\pi}{2}\):
\[ -\cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right) = -0 - 1 = -1 \]
2. At \(x = \frac{\pi}{4}\):
\[ -\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} \]

### Step 7: Combine the results
Now subtract the two results:
\[ -1 - (-\sqrt{2}) = -1 + \sqrt{2} = \sqrt{2} - 1 \]

### Final Answer
Thus, the value of the integral \( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 - \sin 2x} \, dx \) is:
\[ \sqrt{2} - 1 \]