If the integral `I=int_(0)^(pi)=(sec^(-1)(secx))/(1+tan^(8)x)dx, AA x ne (pi)/(2)`, then the value of `[I]` is equal to (where `[.]` is the greatest integer function)

Answer

To solve the integral

\[ I = \int_{0}^{\pi} \frac{\sec^{-1}(\sec x)}{1 + \tan^8 x} \, dx \]

we will use the property of definite integrals and some transformations.

### Step 1: Recognize the Integral and Apply the Property

We can use the property of definite integrals:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

For our case, \( a = 0 \) and \( b = \pi \). Thus, we have:

\[ I = \int_{0}^{\pi} \frac{\sec^{-1}(\sec(\pi - x))}{1 + \tan^8(\pi - x)} \, dx \]

### Step 2: Simplify the Expression

Using the identities:

- \(\sec(\pi - x) = -\sec x\)
- \(\tan(\pi - x) = -\tan x\)

We can rewrite the integral:

\[ I = \int_{0}^{\pi} \frac{\sec^{-1}(-\sec x)}{1 + \tan^8(-x)} \, dx \]

Since \(\sec^{-1}(-\sec x) = \pi - \sec^{-1}(\sec x)\), we have:

\[ I = \int_{0}^{\pi} \frac{\pi - \sec^{-1}(\sec x)}{1 + \tan^8 x} \, dx \]

### Step 3: Combine the Two Integrals

Now we can add the two expressions for \(I\):

\[ 2I = \int_{0}^{\pi} \left( \frac{\sec^{-1}(\sec x)}{1 + \tan^8 x} + \frac{\pi - \sec^{-1}(\sec x)}{1 + \tan^8 x} \right) \, dx \]

This simplifies to:

\[ 2I = \int_{0}^{\pi} \frac{\pi}{1 + \tan^8 x} \, dx \]

### Step 4: Solve for \(I\)

Thus, we can express \(I\) as:

\[ I = \frac{1}{2} \int_{0}^{\pi} \frac{\pi}{1 + \tan^8 x} \, dx \]

### Step 5: Evaluate the Integral

Now, we need to evaluate:

\[ \int_{0}^{\pi} \frac{1}{1 + \tan^8 x} \, dx \]

This integral can be computed using symmetry and known results, but for the sake of this problem, we can use the fact that:

\[ \int_{0}^{\pi} \frac{1}{1 + \tan^8 x} \, dx = \frac{\pi}{2} \]

Thus,

\[ I = \frac{1}{2} \cdot \frac{\pi^2}{2} = \frac{\pi^2}{4} \]

### Step 6: Find the Greatest Integer Function

Now, we calculate:

\[ \lfloor I \rfloor = \lfloor \frac{\pi^2}{4} \rfloor \]

Using \(\pi \approx 3.14\), we find:

\[ \pi^2 \approx 9.86 \implies \frac{\pi^2}{4} \approx 2.465 \]

Thus,

\[ \lfloor I \rfloor = 2 \]

### Final Answer

The value of \([I]\) is:

\[ \boxed{2} \]