`intsqrt(1-cos 2x) dx`

Answer

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

### Step 1: Simplify the expression under the square root
We start with the expression \( 1 - \cos 2x \). We can use the trigonometric identity:

\[ \cos 2x = 1 - 2\sin^2 x \]

Thus,

\[ 1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x \]

### Step 2: Substitute the simplified expression into the integral
Now we can substitute this back into our integral:

\[ \int \sqrt{1 - \cos 2x} \, dx = \int \sqrt{2\sin^2 x} \, dx \]

### Step 3: Simplify the square root
The square root of \( 2\sin^2 x \) can be simplified as follows:

\[ \sqrt{2\sin^2 x} = \sqrt{2} \cdot \sqrt{\sin^2 x} = \sqrt{2} \cdot |\sin x| \]

Assuming \( x \) is in a range where \( \sin x \) is non-negative (for example, \( 0 \leq x \leq \pi \)), we can write:

\[ \sqrt{2\sin^2 x} = \sqrt{2} \sin x \]

### Step 4: Substitute back into the integral
Now we substitute this back into the integral:

\[ \int \sqrt{2} \sin x \, dx \]

### Step 5: Factor out the constant
We can factor out \( \sqrt{2} \):

\[ \sqrt{2} \int \sin x \, dx \]

### Step 6: Integrate \( \sin x \)
The integral of \( \sin x \) is:

\[ \int \sin x \, dx = -\cos x \]

### Step 7: Combine the results
Putting it all together, we have:

\[ \sqrt{2} \int \sin x \, dx = \sqrt{2} (-\cos x) + C = -\sqrt{2} \cos x + C \]

### Final Result
Thus, the final result of the integral is:

\[ \int \sqrt{1 - \cos 2x} \, dx = -\sqrt{2} \cos x + C \]