Let `f(x)={{:(((1-cosx)/((2pi-x)^(2)))((sin^(2)x)/(log(1+4pi^(2)-4pix+x^(2)))),:,xne2pi),(" "lambda,:,x=2pi):}`
is continuous at `x=2pi`, then the value of `lambda` is equal to

Answer

To find the value of \(\lambda\) such that the function \(f(x)\) is continuous at \(x = 2\pi\), we need to evaluate the limit of \(f(x)\) as \(x\) approaches \(2\pi\) and set it equal to \(\lambda\).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \frac{(1 - \cos x)}{(2\pi - x)^2} \cdot \frac{\sin^2 x}{\log(1 + 4\pi^2 - 4\pi x + x^2)} \quad \text{for } x \neq 2\pi \]
and
\[ f(2\pi) = \lambda. \]

2. **Find the limit as \(x\) approaches \(2\pi\)**:
We need to compute:
\[ \lim_{x \to 2\pi} f(x). \]

3. **Substitute \(x = 2\pi + t\)**:
As \(x\) approaches \(2\pi\), let \(t = x - 2\pi\), so \(x = 2\pi + t\) and \(t \to 0\) as \(x \to 2\pi\).

4. **Rewrite the limit**:
The limit becomes:
\[ \lim_{t \to 0} f(2\pi + t) = \lim_{t \to 0} \frac{(1 - \cos(2\pi + t))}{(2\pi - (2\pi + t))^2} \cdot \frac{\sin^2(2\pi + t)}{\log(1 + 4\pi^2 - 4\pi(2\pi + t) + (2\pi + t)^2)}. \]

5. **Simplify the terms**:
- Since \(\cos(2\pi + t) = \cos t\) and \(\sin(2\pi + t) = \sin t\), we have:
\[ 1 - \cos(2\pi + t) = 1 - \cos t. \]
- The denominator becomes:
\[ (2\pi - (2\pi + t))^2 = (-t)^2 = t^2. \]
- The logarithm simplifies as follows:
\[ 1 + 4\pi^2 - 4\pi(2\pi + t) + (2\pi + t)^2 = 1 + 4\pi^2 - 8\pi^2 - 4\pi t + 4\pi^2 + 4\pi t + t^2 = 1 + t^2. \]

6. **Combine the expressions**:
Thus, we have:
\[ f(2\pi + t) = \frac{(1 - \cos t)}{t^2} \cdot \frac{\sin^2 t}{\log(1 + t^2)}. \]

7. **Evaluate the limit**:
- Using the limit \(\lim_{t \to 0} \frac{1 - \cos t}{t^2} = \frac{1}{2}\) and \(\lim_{t \to 0} \frac{\sin^2 t}{t^2} = 1\), we need to evaluate:
\[ \lim_{t \to 0} \frac{\sin^2 t}{\log(1 + t^2)}. \]
- This is of the form \(\frac{0}{0}\), so we apply L'Hôpital's rule:
\[ \lim_{t \to 0} \frac{2\sin t \cos t}{\frac{2t}{1 + t^2}} = \lim_{t \to 0} \frac{\sin(2t)}{t} = 2. \]

8. **Final limit calculation**:
Therefore, we find:
\[ \lim_{t \to 0} f(2\pi + t) = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}. \]

9. **Set the limit equal to \(\lambda\)**:
Since \(f(x)\) is continuous at \(x = 2\pi\):
\[ \lambda = \frac{1}{2}. \]

### Conclusion:
The value of \(\lambda\) is \(\frac{1}{2}\).