Evaluate: `int_0^(100pi)sqrt((1-cos2x))dxdot`

Explanation

To evaluate the integral

\[ I = \int_0^{100\pi} \sqrt{1 - \cos 2x} \, dx, \]

we will follow these steps:

### Step 1: Simplify the integrand using a trigonometric identity

We know from trigonometric identities that

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

Thus, we can rewrite the integral as:

\[ I = \int_0^{100\pi} \sqrt{2 \sin^2 x} \, dx. \]

### Step 2: Factor out constants from the square root

The square root can be simplified:

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

So, we can express the integral as:

\[ I = \sqrt{2} \int_0^{100\pi} |\sin x| \, dx. \]

### Step 3: Analyze the periodicity of the sine function

The function \(|\sin x|\) is periodic with a period of \(\pi\). Therefore, we can break the integral into intervals of \(\pi\):

\[ I = \sqrt{2} \int_0^{100\pi} |\sin x| \, dx = \sqrt{2} \cdot \left(\int_0^{\pi} |\sin x| \, dx\right) \cdot \frac{100\pi}{\pi}. \]

### Step 4: Calculate the integral over one period

Now we need to calculate

\[ \int_0^{\pi} |\sin x| \, dx. \]

Since \(\sin x\) is non-negative on the interval \([0, \pi]\), we have:

\[ \int_0^{\pi} |\sin x| \, dx = \int_0^{\pi} \sin x \, dx. \]

The integral of \(\sin x\) is:

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

Evaluating this from \(0\) to \(\pi\):

\[ \int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2. \]

### Step 5: Substitute back into the equation for \(I\)

Now substituting back, we have:

\[ I = \sqrt{2} \cdot 2 \cdot 100 = 200\sqrt{2}. \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{200\sqrt{2}}. \]