`int_(0)^(pi//6) sqrt(1-sin 2x) dx`

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{6}} \sqrt{1 - \sin 2x} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
Using the double angle identity for sine, we know that:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the integrand:
\[ 1 - \sin 2x = 1 - 2 \sin x \cos x \]
We can also use the identity \( \cos^2 x + \sin^2 x = 1 \) to express this in a different form:
\[ 1 - 2 \sin x \cos x = (\sin x - \cos x)^2 \]
So, we have:
\[ \sqrt{1 - \sin 2x} = \sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x| \]

### Step 2: Determine the sign of the expression
For \( x \) in the interval \( [0, \frac{\pi}{6}] \):
- At \( x = 0 \), \( \sin 0 - \cos 0 = 0 - 1 = -1 \)
- At \( x = \frac{\pi}{6} \), \( \sin \frac{\pi}{6} - \cos \frac{\pi}{6} = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1 - \sqrt{3}}{2} \) which is negative.

Thus, \( \sin x - \cos x \) is negative in the interval \( [0, \frac{\pi}{6}] \). Therefore:
\[ |\sin x - \cos x| = -(\sin x - \cos x) = \cos x - \sin x \]

### Step 3: Rewrite the integral
Now we can rewrite the integral:
\[ \int_{0}^{\frac{\pi}{6}} \sqrt{1 - \sin 2x} \, dx = \int_{0}^{\frac{\pi}{6}} (\cos x - \sin x) \, dx \]

### Step 4: Integrate the expression
Now we can integrate:
\[ \int (\cos x - \sin x) \, dx = \int \cos x \, dx - \int \sin x \, dx = \sin x + \cos x + C \]

### Step 5: Apply the limits
Now we evaluate the definite integral from \( 0 \) to \( \frac{\pi}{6} \):
\[ \left[ \sin x + \cos x \right]_{0}^{\frac{\pi}{6}} = \left( \sin \frac{\pi}{6} + \cos \frac{\pi}{6} \right) - \left( \sin 0 + \cos 0 \right) \]
Calculating the values:
- \( \sin \frac{\pi}{6} = \frac{1}{2} \)
- \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)
- \( \sin 0 = 0 \)
- \( \cos 0 = 1 \)

Thus:
\[ \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) - (0 + 1) = \left( \frac{1 + \sqrt{3}}{2} \right) - 1 = \frac{1 + \sqrt{3} - 2}{2} = \frac{\sqrt{3} - 1}{2} \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{\frac{\pi}{6}} \sqrt{1 - \sin 2x} \, dx = \frac{\sqrt{3} - 1}{2} \]