`int_(1)^(10pi)([sec^(-1)x]+[cot^(-1)x])dx`, where [.] denotes the greatest integer function, is equal to:

Options

A. `10 pi` - sec1
B. `10pi +` secl
C. `10pi - secl + cotl`
D. secl + cotl

Answer

To solve the integral \( I = \int_{1}^{10\pi} \left( \lfloor \sec^{-1} x \rfloor + \lfloor \cot^{-1} x \rfloor \right) dx \), where \( \lfloor . \rfloor \) denotes the greatest integer function, we can break it down into two parts:

1. **Evaluate \( \int_{1}^{10\pi} \lfloor \sec^{-1} x \rfloor dx \)**
2. **Evaluate \( \int_{1}^{10\pi} \lfloor \cot^{-1} x \rfloor dx \)**

### Step 1: Evaluate \( \int_{1}^{10\pi} \lfloor \sec^{-1} x \rfloor dx \)

**Finding the range of \( \sec^{-1} x \):**
- The function \( \sec^{-1} x \) is defined for \( x \geq 1 \).
- As \( x \) increases from 1 to \( 10\pi \), \( \sec^{-1} x \) increases from \( \sec^{-1}(1) = 0 \) to \( \sec^{-1}(10\pi) \).

**Finding the value of \( \sec^{-1}(10\pi) \):**
- Since \( 10\pi \) is a large number, we can approximate \( \sec^{-1}(10\pi) \) as a large angle, which will be greater than \( \frac{\pi}{2} \).

**Identifying the greatest integer values:**
- For \( x \) in the interval \( [1, \sec(1)] \), \( \lfloor \sec^{-1} x \rfloor = 0 \).
- For \( x \) in the interval \( [\sec(1), 10\pi] \), \( \lfloor \sec^{-1} x \rfloor = 1 \).

**Calculating the integral:**
- The integral can be split as follows:
\[ \int_{1}^{10\pi} \lfloor \sec^{-1} x \rfloor dx = \int_{1}^{\sec(1)} 0 \, dx + \int_{\sec(1)}^{10\pi} 1 \, dx \]
- The first integral evaluates to \( 0 \).
- The second integral evaluates to:
\[ \int_{\sec(1)}^{10\pi} 1 \, dx = 10\pi - \sec(1) \]

### Step 2: Evaluate \( \int_{1}^{10\pi} \lfloor \cot^{-1} x \rfloor dx \)

**Finding the range of \( \cot^{-1} x \):**
- The function \( \cot^{-1} x \) is defined for all \( x > 0 \).
- As \( x \) increases from 1 to \( 10\pi \), \( \cot^{-1} x \) decreases from \( \cot^{-1}(1) = \frac{\pi}{4} \) to \( \cot^{-1}(10\pi) \) which approaches \( 0 \).

**Identifying the greatest integer values:**
- For \( x \) in the interval \( [1, 10\pi] \), \( \lfloor \cot^{-1} x \rfloor \) will be \( 0 \) for all \( x \) since \( \cot^{-1}(1) = \frac{\pi}{4} \approx 0.785 \) and \( \cot^{-1}(10\pi) \) is less than \( 1 \).

**Calculating the integral:**
- Thus, we have:
\[ \int_{1}^{10\pi} \lfloor \cot^{-1} x \rfloor dx = \int_{1}^{10\pi} 0 \, dx = 0 \]

### Final Calculation
Combining both parts:
\[ I = \int_{1}^{10\pi} \lfloor \sec^{-1} x \rfloor dx + \int_{1}^{10\pi} \lfloor \cot^{-1} x \rfloor dx = (10\pi - \sec(1)) + 0 = 10\pi - \sec(1) \]

### Conclusion
The value of the integral is:
\[ \int_{1}^{10\pi} \left( \lfloor \sec^{-1} x \rfloor + \lfloor \cot^{-1} x \rfloor \right) dx = 10\pi - \sec(1) \]