Solve the following integration
`int sqrt(1-sin 2x)dx`

Answer

To solve the integral \( \int \sqrt{1 - \sin 2x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ \int \sqrt{1 - \sin 2x} \, dx \]
We know that \( \sin 2x = 2 \sin x \cos x \). Therefore, we can rewrite \( 1 - \sin 2x \) as:
\[ 1 - \sin 2x = 1 - 2 \sin x \cos x \]

### Step 2: Use Pythagorean identity
Next, we can use the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to express \( 1 - 2 \sin x \cos x \):
\[ 1 - 2 \sin x \cos x = \cos^2 x + \sin^2 x - 2 \sin x \cos x \]
This can be rearranged as:
\[ = (\cos x - \sin x)^2 \]

### Step 3: Substitute back into the integral
Now we substitute this back into our integral:
\[ \int \sqrt{1 - \sin 2x} \, dx = \int \sqrt{(\cos x - \sin x)^2} \, dx \]
Since the square root of a square is the absolute value, we have:
\[ = \int |\cos x - \sin x| \, dx \]

### Step 4: Determine the sign of the expression
To proceed, we need to determine when \( \cos x - \sin x \) is positive or negative. The expression \( \cos x - \sin x = 0 \) when \( \tan x = 1 \), which occurs at \( x = \frac{\pi}{4} + n\pi \) for integer \( n \).

For \( 0 \leq x < \frac{\pi}{4} \), \( \cos x - \sin x \) is positive, and for \( \frac{\pi}{4} < x < \frac{\pi}{2} \), it is negative. Thus, we can split our integral based on these intervals.

### Step 5: Solve the integral
For the interval \( [0, \frac{\pi}{4}] \):
\[ \int (\cos x - \sin x) \, dx = \int \cos x \, dx - \int \sin x \, dx \]
Calculating these gives:
\[ = \sin x + \cos x + C \]

For the interval \( [\frac{\pi}{4}, \frac{\pi}{2}] \):
\[ \int -(\cos x - \sin x) \, dx = -\int (\cos x - \sin x) \, dx = -(\sin x + \cos x) + C \]

### Final Result
Thus, the final result of the integral \( \int \sqrt{1 - \sin 2x} \, dx \) is:
\[ \int \sqrt{1 - \sin 2x} \, dx = \begin{cases} \sin x + \cos x + C & \text{for } 0 \leq x < \frac{\pi}{4} \\ -\sin x - \cos x + C & \text{for } \frac{\pi}{4} < x < \frac{\pi}{2} \end{cases} \]