The value of the integral
`int_(0)^(400pi) sqrt(1-cos2x)dx`, is

Options

A. `200sqrt(2)`
B. `400sqrt(2)`
C. `800sqrt(2)`
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{400\pi} \sqrt{1 - \cos 2x} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
Using the trigonometric identity, we know that:
\[ \cos 2x = 1 - 2\sin^2 x \]
Thus, we can rewrite \( 1 - \cos 2x \):
\[ 1 - \cos 2x = 2\sin^2 x \]
So, we can express the integral as:
\[ I = \int_{0}^{400\pi} \sqrt{2\sin^2 x} \, dx \]

### Step 2: Factor out constants
The square root can be simplified:
\[ \sqrt{2\sin^2 x} = \sqrt{2} \cdot |\sin x| \]
Therefore, the integral becomes:
\[ I = \sqrt{2} \int_{0}^{400\pi} |\sin x| \, dx \]

### Step 3: Analyze the periodicity of the function
The function \( |\sin x| \) has a period of \( \pi \). Thus, we can break the integral from \( 0 \) to \( 400\pi \) into intervals of \( \pi \):
\[ I = \sqrt{2} \cdot \left( \int_{0}^{\pi} |\sin x| \, dx \right) \cdot \text{number of periods} \]
The number of periods in \( [0, 400\pi] \) is:
\[ \frac{400\pi}{\pi} = 400 \]
So we have:
\[ I = 400 \sqrt{2} \int_{0}^{\pi} |\sin x| \, dx \]

### Step 4: Calculate the integral over one period
Since \( \sin x \) is non-negative in the interval \( [0, \pi] \), we have:
\[ |\sin x| = \sin x \quad \text{for } x \in [0, \pi] \]
Thus:
\[ \int_{0}^{\pi} |\sin x| \, dx = \int_{0}^{\pi} \sin x \, dx \]
The integral of \( \sin x \) is:
\[ -\cos x \bigg|_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

### Step 5: Substitute back into the equation
Now substituting back into our expression for \( I \):
\[ I = 400 \sqrt{2} \cdot 2 = 800 \sqrt{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{800\sqrt{2}} \]