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

Options

A. (a)`(sec1)-10pi`
B. (b)`10pi-sec1`
C. (c)`pi-sec 1`
D. (d)None of these

Answer

To solve the integral \( \int_{1}^{10\pi} [\sec^{-1} x] \, dx \) where \( [.] \) denotes the greatest integer function, we will follow these steps:

### Step 1: Understand the function \( \sec^{-1} x \)
The function \( \sec^{-1} x \) is defined for \( x \geq 1 \) and \( x \leq -1 \). The range of \( \sec^{-1} x \) is \( [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi] \).

### Step 2: Identify the intervals for integration
Since we are integrating from \( 1 \) to \( 10\pi \), we need to determine the values of \( \sec^{-1} x \) within this interval.

- For \( x = 1 \), \( \sec^{-1}(1) = 0 \).
- For \( x = 2 \), \( \sec^{-1}(2) = \frac{\pi}{3} \).
- For \( x = 3 \), \( \sec^{-1}(3) \) is approximately \( 1.249 \) radians.
- Continuing this way, we find that \( \sec^{-1}(x) \) will keep increasing as \( x \) increases.

### Step 3: Determine the greatest integer function values
We need to find the greatest integer values of \( \sec^{-1} x \) for \( x \) in the interval \( [1, 10\pi] \).

- For \( 1 \leq x < 2 \): \( [\sec^{-1} x] = 0 \)
- For \( 2 \leq x < 3 \): \( [\sec^{-1} x] = 1 \)
- For \( 3 \leq x < 4 \): \( [\sec^{-1} x] = 1 \)
- For \( 4 \leq x < 5 \): \( [\sec^{-1} x] = 1 \)
- For \( 5 \leq x < 6 \): \( [\sec^{-1} x] = 1 \)
- For \( 6 \leq x < 7 \): \( [\sec^{-1} x] = 1 \)
- For \( 7 \leq x < 8 \): \( [\sec^{-1} x] = 1 \)
- For \( 8 \leq x < 9 \): \( [\sec^{-1} x] = 1 \)
- For \( 9 \leq x < 10 \): \( [\sec^{-1} x] = 1 \)
- For \( 10 \leq x < 10\pi \): \( [\sec^{-1} x] = 2 \) (since \( \sec^{-1}(10) \) is greater than \( 2 \) but less than \( \frac{3\pi}{2} \))

### Step 4: Set up the integral based on intervals
Now we can break the integral into parts based on the intervals we identified:

\[ \int_{1}^{10\pi} [\sec^{-1} x] \, dx = \int_{1}^{2} 0 \, dx + \int_{2}^{10} 1 \, dx + \int_{10}^{10\pi} 2 \, dx \]

### Step 5: Calculate each integral
1. \( \int_{1}^{2} 0 \, dx = 0 \)
2. \( \int_{2}^{10} 1 \, dx = [x]_{2}^{10} = 10 - 2 = 8 \)
3. \( \int_{10}^{10\pi} 2 \, dx = 2[x]_{10}^{10\pi} = 2(10\pi - 10) = 20\pi - 20 \)

### Step 6: Combine the results
Now we sum the results of the integrals:

\[ \int_{1}^{10\pi} [\sec^{-1} x] \, dx = 0 + 8 + (20\pi - 20) = 20\pi - 12 \]

### Final Answer
Thus, the value of the integral \( \int_{1}^{10\pi} [\sec^{-1} x] \, dx \) is \( 20\pi - 12 \).

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