If the middle term in the binomial expansion of `((1)/(x) + x sin x )^(10) ` is `( 63)/( 8)` , then the value of `6sin^(2) x +sin x -2` is

Answer

To solve the given problem step by step, we will follow the process of finding the middle term in the binomial expansion and then substituting the value of \( \sin x \) to find the desired expression.

### Step 1: Identify the Binomial Expansion
We start with the expression:
\[ \left( \frac{1}{x} + x \sin x \right)^{10} \]
This is a binomial expression where \( a = \frac{1}{x} \) and \( b = x \sin x \).

### Step 2: Determine the Middle Term
For a binomial expansion of the form \( (a + b)^n \), the middle term (when \( n \) is even) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
where \( r = \frac{n}{2} \).

Here, \( n = 10 \), so the middle term is:
\[ T_{6} = \binom{10}{5} \left( \frac{1}{x} \right)^{10-5} (x \sin x)^5 \]

### Step 3: Simplify the Middle Term
Calculating \( T_6 \):
\[ T_6 = \binom{10}{5} \left( \frac{1}{x} \right)^5 (x \sin x)^5 \]
\[ = \binom{10}{5} \cdot \frac{1}{x^5} \cdot x^5 \cdot (\sin x)^5 \]
\[ = \binom{10}{5} \cdot (\sin x)^5 \]

### Step 4: Calculate \( \binom{10}{5} \)
The value of \( \binom{10}{5} \) is calculated as follows:
\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Step 5: Set Up the Equation
We know from the problem statement that:
\[ 252 \cdot (\sin x)^5 = \frac{63}{8} \]

### Step 6: Solve for \( \sin x \)
Now we can solve for \( \sin x \):
\[ (\sin x)^5 = \frac{63}{8 \cdot 252} \]
Calculating \( 8 \cdot 252 = 2016 \):
\[ (\sin x)^5 = \frac{63}{2016} \]
Simplifying \( \frac{63}{2016} \):
\[ = \frac{1}{32} \quad \text{(since } 2016 = 63 \times 32\text{)} \]
Thus, we have:
\[ \sin x = \left( \frac{1}{2} \right) = \frac{1}{2} \]

### Step 7: Substitute \( \sin x \) into the Expression
We need to find the value of:
\[ 6 \sin^2 x + \sin x - 2 \]
Substituting \( \sin x = \frac{1}{2} \):
\[ 6 \left( \frac{1}{2} \right)^2 + \frac{1}{2} - 2 \]
Calculating \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \):
\[ = 6 \cdot \frac{1}{4} + \frac{1}{2} - 2 \]
\[ = \frac{6}{4} + \frac{2}{4} - \frac{8}{4} \]
\[ = \frac{6 + 2 - 8}{4} = \frac{0}{4} = 0 \]

### Final Answer
Thus, the value of \( 6 \sin^2 x + \sin x - 2 \) is:
\[ \boxed{0} \]