`intsqrt(1-cos 2x) dx`

Answer

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

### Step 1: Use the trigonometric identity
We know from trigonometric identities that:
\[ 1 - \cos 2x = 2 \sin^2 x \]
Thus, we can rewrite the integral as:
\[ \int \sqrt{1 - \cos 2x} \, dx = \int \sqrt{2 \sin^2 x} \, dx \]

### Step 2: Simplify the square root
The square root can be simplified:
\[ \sqrt{2 \sin^2 x} = \sqrt{2} \cdot \sin x \]
So, the integral now becomes:
\[ \int \sqrt{2} \cdot \sin x \, dx \]

### Step 3: Factor out the constant
Since \(\sqrt{2}\) is a constant, we can factor it out of the integral:
\[ \sqrt{2} \int \sin x \, dx \]

### Step 4: Integrate \(\sin x\)
The integral of \(\sin x\) is:
\[ \int \sin x \, dx = -\cos x \]
Thus, we have:
\[ \sqrt{2} \cdot (-\cos x) = -\sqrt{2} \cos x \]

### Step 5: Add the constant of integration
Finally, we add the constant of integration \(C\):
\[ -\sqrt{2} \cos x + C \]

### Final Answer
The final result of the integral is:
\[ \int \sqrt{1 - \cos 2x} \, dx = -\sqrt{2} \cos x + C \]