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

Answer

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

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

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

### Step 3: Use the identity for \( \sin^2 x \)
We can use the identity:
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
Substituting this into the integral gives:
\[ \sqrt{2} \int \sin^2 x \, dx = \sqrt{2} \int \frac{1 - \cos 2x}{2} \, dx \]

### Step 4: Simplify the integral
We can factor out the constant:
\[ = \frac{\sqrt{2}}{2} \int (1 - \cos 2x) \, dx \]

### Step 5: Integrate the terms
Now we can integrate each term:
\[ \int (1 - \cos 2x) \, dx = \int 1 \, dx - \int \cos 2x \, dx \]
The integrals evaluate to:
\[ = x - \frac{\sin 2x}{2} + C \]

### Step 6: Combine results
Substituting back, we have:
\[ \frac{\sqrt{2}}{2} \left( x - \frac{\sin 2x}{2} \right) + C \]
This simplifies to:
\[ \frac{\sqrt{2}}{2} x - \frac{\sqrt{2}}{4} \sin 2x + C \]

### Final Answer
Thus, the final answer to the integral is:
\[ \int \sin x \sqrt{1 - \cos 2x} \, dx = \frac{\sqrt{2}}{2} x - \frac{\sqrt{2}}{4} \sin 2x + C \]