`int_(-4)^(4) (sin^(-1)( sinx)+ cos^(-1)(cos x))/((1+x^(2))(1+[(x^(2))/(17)])) dx= log (((1+pi^(2)))/sqrt(a)) `
`bpi" tan"^(-1)((c-pi)/(1+c pi))` (where, [.] denotes greatest inteher function), then the number of ways in which `a-(2b+c)` distinct object can distributed among `(a-5)/(c)` persons equally , is

Options

A. `(9!)/((3!)^(3))`
B. `(12!)/((140^(3))`
C. `(15!)/((5!)^(3))`
D. `(10!)/((6!)^(3))`

Answer

To solve the given integral and find the values of \( a \), \( b \), and \( c \), we will follow these steps:

### Step 1: Evaluate the Integral
We need to evaluate the integral:
\[ I = \int_{-4}^{4} \frac{\sin^{-1}(\sin x) + \cos^{-1}(\cos x)}{(1+x^2)\left(1+\left\lfloor \frac{x^2}{17} \right\rfloor\right)} \, dx \]

### Step 2: Analyze the Greatest Integer Function
Since \( x \) varies from \(-4\) to \(4\), \( x^2 \) varies from \(0\) to \(16\). Therefore, \( \left\lfloor \frac{x^2}{17} \right\rfloor = 0 \) for all \( x \) in this range. Thus, the integral simplifies to:
\[ I = \int_{-4}^{4} \frac{\sin^{-1}(\sin x) + \cos^{-1}(\cos x)}{1+x^2} \, dx \]

### Step 3: Split the Integral
We can split the integral into two parts:
\[ I = \int_{-4}^{4} \frac{\sin^{-1}(\sin x)}{1+x^2} \, dx + \int_{-4}^{4} \frac{\cos^{-1}(\cos x)}{1+x^2} \, dx \]

### Step 4: Evaluate Each Integral
1. **For \( \sin^{-1}(\sin x) \)**:
- This function is odd, so:
\[ \int_{-4}^{4} \frac{\sin^{-1}(\sin x)}{1+x^2} \, dx = 0 \]

2. **For \( \cos^{-1}(\cos x) \)**:
- We need to evaluate:
\[ \int_{-4}^{4} \frac{\cos^{-1}(\cos x)}{1+x^2} \, dx \]
- Note that \( \cos^{-1}(\cos x) \) is periodic and symmetric. We can break it into two intervals:
\[ \int_{0}^{\pi} \frac{x}{1+x^2} \, dx + \int_{\pi}^{4} \frac{2\pi - x}{1+x^2} \, dx \]

### Step 5: Evaluate the First Integral
For \( \int_{0}^{\pi} \frac{x}{1+x^2} \, dx \):
Using integration by parts, let \( u = x \) and \( dv = \frac{dx}{1+x^2} \):
\[ du = dx, \quad v = \tan^{-1}(x) \]
Thus,
\[ \int_{0}^{\pi} \frac{x}{1+x^2} \, dx = \left[ x \tan^{-1}(x) \right]_{0}^{\pi} - \int_{0}^{\pi} \tan^{-1}(x) \, dx \]

### Step 6: Evaluate the Second Integral
For \( \int_{\pi}^{4} \frac{2\pi - x}{1+x^2} \, dx \):
This can be evaluated similarly, leading to:
\[ \int_{\pi}^{4} \frac{2\pi - x}{1+x^2} \, dx = 2\pi \tan^{-1}(4) - \int_{\pi}^{4} \tan^{-1}(x) \, dx \]

### Step 7: Combine Results
Combine the results from both integrals to find \( I \).

### Step 8: Comparing with Given Expression
After evaluating \( I \), we compare it with:
\[ \log\left(\frac{1+\pi^2}{\sqrt{a}}\right) + b\pi \tan^{-1}\left(\frac{c-\pi}{1+c\pi}\right) \]
From this, we can determine \( a \), \( b \), and \( c \).

### Step 9: Calculate \( a - 2b + c \)
Using the values of \( a \), \( b \), and \( c \) obtained from the comparison, calculate \( a - 2b + c \).

### Step 10: Distributing Objects
Finally, we need to find the number of ways to distribute \( a - 2b + c \) distinct objects among \( \frac{a-5}{c} \) persons equally:
\[ \text{Number of ways} = \frac{(a - 2b + c)!}{\left(\frac{a-5}{c}\right)!^{\frac{a-2b+c}{\frac{a-5}{c}}}} \]