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

\[ f(x) = \begin{cases} \frac{(1 - \cos x)}{(2\pi - x)^2} \cdot \frac{\sin^2 x}{\log(1 + 4\pi^2 - 4\pi x + x^2)} & \text{if } x \neq 2\pi \\ \lambda & \text{if } x = 2\pi \end{cases} \]

is continuous at \( x = 2\pi \), we need to ensure that

\[ \lim_{x \to 2\pi} f(x) = f(2\pi) = \lambda. \]

### Step 1: Calculate the limit as \( x \to 2\pi \)

We need to evaluate:

\[ \lim_{x \to 2\pi} \frac{(1 - \cos x)}{(2\pi - x)^2} \cdot \frac{\sin^2 x}{\log(1 + 4\pi^2 - 4\pi x + x^2)}. \]

### Step 2: Substitute \( x = 2\pi + t \)

Let \( x = 2\pi + t \) where \( t \to 0 \) as \( x \to 2\pi \). Then, we have:

\[ \lim_{t \to 0} \frac{(1 - \cos(2\pi + t))}{(2\pi - (2\pi + t))^2} \cdot \frac{\sin^2(2\pi + t)}{\log(1 + 4\pi^2 - 4\pi(2\pi + t) + (2\pi + t)^2)}. \]

### Step 3: Simplify the expressions

Using the identities \( \cos(2\pi + t) = \cos t \) and \( \sin(2\pi + t) = \sin t \), we can rewrite the limit as:

\[ \lim_{t \to 0} \frac{(1 - \cos t)}{t^2} \cdot \frac{\sin^2 t}{\log(1 + 4\pi^2 - 8\pi^2 - 4\pi t + 4\pi^2 + 4\pi t + t^2)}. \]

This simplifies to:

\[ \lim_{t \to 0} \frac{(1 - \cos t)}{t^2} \cdot \frac{\sin^2 t}{\log(1 + t^2)}. \]

### Step 4: Use standard limits

We know the following limits:

1. \( \lim_{t \to 0} \frac{1 - \cos t}{t^2} = \frac{1}{2} \)
2. \( \lim_{t \to 0} \frac{\sin^2 t}{t^2} = 1 \)
3. \( \lim_{t \to 0} \frac{t^2}{\log(1 + t^2)} = 1 \)

Thus, we can evaluate the limit:

\[ \lim_{t \to 0} \frac{(1 - \cos t)}{t^2} \cdot \frac{\sin^2 t}{\log(1 + t^2)} = \left(\frac{1}{2}\right) \cdot 1 \cdot 1 = \frac{1}{2}. \]

### Step 5: Set the limit equal to \( \lambda \)

Since the function is continuous at \( x = 2\pi \), we set:

\[ \frac{1}{2} = \lambda. \]

### Conclusion

Thus, the value of \( \lambda \) is

\[ \boxed{\frac{1}{2}}. \]