`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: Rewrite the integrand
We start by rewriting \( 1 - \sin 2x \). We know that:
\[ 1 = \sin^2 x + \cos^2 x \]
And we can express \( \sin 2x \) as:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the expression under the square root:
\[ 1 - \sin 2x = 1 - 2 \sin x \cos x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\sin x - \cos x)^2 \]

### Step 2: Substitute into the integral
Now we substitute this back into the integral:
\[ \int_0^{\frac{\pi}{6}} \sqrt{1 - \sin 2x} \, dx = \int_0^{\frac{\pi}{6}} \sqrt{(\sin x - \cos x)^2} \, dx \]
Since \( \sin x - \cos x \) is non-negative in the interval \( [0, \frac{\pi}{6}] \), we can simplify:
\[ \sqrt{(\sin x - \cos x)^2} = \sin x - \cos x \]
Thus, the integral becomes:
\[ \int_0^{\frac{\pi}{6}} (\sin x - \cos x) \, dx \]

### Step 3: Integrate
Now we can integrate:
\[ \int (\sin x - \cos x) \, dx = -\cos x - \sin x + C \]
Now we evaluate the definite integral from \( 0 \) to \( \frac{\pi}{6} \):
\[ \left[-\cos x - \sin x\right]_0^{\frac{\pi}{6}} = \left[-\cos\left(\frac{\pi}{6}\right) - \sin\left(\frac{\pi}{6}\right)\right] - \left[-\cos(0) - \sin(0)\right] \]

### Step 4: Calculate the limits
Now we calculate the values:
- At \( x = \frac{\pi}{6} \):
\[ -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}, \quad -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2} \]
So:
\[ -\cos\left(\frac{\pi}{6}\right) - \sin\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} - \frac{1}{2} = -\frac{\sqrt{3} + 1}{2} \]

- At \( x = 0 \):
\[ -\cos(0) = -1, \quad -\sin(0) = 0 \]
So:
\[ -\cos(0) - \sin(0) = -1 \]

### Step 5: Final calculation
Putting it all together:
\[ \left[-\cos\left(\frac{\pi}{6}\right) - \sin\left(\frac{\pi}{6}\right)\right] - \left[-\cos(0) - \sin(0)\right] = \left(-\frac{\sqrt{3} + 1}{2}\right) - (-1) \]
This simplifies to:
\[ -\frac{\sqrt{3} + 1}{2} + 1 = 1 - \frac{\sqrt{3} + 1}{2} = \frac{2 - (\sqrt{3} + 1)}{2} = \frac{1 - \sqrt{3}}{2} \]

### Final Answer
Thus, the value of the integral \( \int_0^{\frac{\pi}{6}} \sqrt{1 - \sin 2x} \, dx \) is:
\[ \frac{1 - \sqrt{3}}{2} \]