Evaluate the following integrals.
` int sin x sqrt( 1 - cos 2x)dx`

Answer

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

### Step 1: Simplify the integrand
We know that \( \cos 2x = 1 - 2\sin^2 x \). Therefore, we can express \( 1 - \cos 2x \) as:
\[ 1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x \]
Thus, we can rewrite the integral as:
\[ I = \int \sin x \sqrt{2 \sin^2 x} \, dx \]

### Step 2: Factor out constants
Since \( \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x| \) and assuming \( \sin x \) is non-negative in the interval we are considering, we can simplify this to:
\[ I = \sqrt{2} \int \sin^2 x \, dx \]

### Step 3: Use the identity for \( \sin^2 x \)
We can use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
\[ I = \sqrt{2} \int \frac{1 - \cos 2x}{2} \, dx \]
This simplifies to:
\[ I = \frac{\sqrt{2}}{2} \int (1 - \cos 2x) \, dx \]

### Step 4: Integrate the expression
Now we can integrate term by term:
\[ I = \frac{\sqrt{2}}{2} \left( \int 1 \, dx - \int \cos 2x \, dx \right) \]
The integrals are:
\[ \int 1 \, dx = x \]
\[ \int \cos 2x \, dx = \frac{\sin 2x}{2} \]
Thus, we have:
\[ I = \frac{\sqrt{2}}{2} \left( x - \frac{\sin 2x}{2} \right) + C \]

### Step 5: Simplify the result
Distributing \( \frac{\sqrt{2}}{2} \):
\[ I = \frac{\sqrt{2}}{2} x - \frac{\sqrt{2}}{4} \sin 2x + C \]

### Final Answer
The final result of the integral is:
\[ I = \frac{\sqrt{2}}{2} x - \frac{\sqrt{2}}{4} \sin 2x + C \]