Let `I = int _(0) ^(pi) x ^(6) (pi-x) ^(8)dx,` then `(pi ^(15))/((""^(15) C _(9))I )=`

Answer

To solve the integral \( I = \int_0^{\pi} x^6 (\pi - x)^8 \, dx \) and find the value of \( \frac{\pi^{15}}{\binom{15}{9} I} \), we will use the integration by parts method repeatedly.

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_0^{\pi} x^6 (\pi - x)^8 \, dx \]

### Step 2: Use integration by parts
Let:
- \( u = x^6 \) → \( du = 6x^5 \, dx \)
- \( dv = (\pi - x)^8 \, dx \) → \( v = -\frac{(\pi - x)^9}{9} \)

Using integration by parts:
\[ I = \left[ -\frac{x^6 (\pi - x)^9}{9} \right]_0^{\pi} + \frac{1}{9} \int_0^{\pi} 6x^5 (\pi - x)^9 \, dx \]

### Step 3: Evaluate the boundary terms
Evaluating the boundary terms:
\[ \left[ -\frac{x^6 (\pi - x)^9}{9} \right]_0^{\pi} = -\frac{\pi^6 \cdot 0^9}{9} + \frac{0^6 \cdot \pi^9}{9} = 0 \]
Thus, we have:
\[ I = \frac{6}{9} \int_0^{\pi} x^5 (\pi - x)^9 \, dx = \frac{2}{3} \int_0^{\pi} x^5 (\pi - x)^9 \, dx \]

### Step 4: Repeat integration by parts
Now, we apply integration by parts again:
Let:
- \( u = x^5 \) → \( du = 5x^4 \, dx \)
- \( dv = (\pi - x)^9 \, dx \) → \( v = -\frac{(\pi - x)^{10}}{10} \)

Then:
\[ \int_0^{\pi} x^5 (\pi - x)^9 \, dx = \left[ -\frac{x^5 (\pi - x)^{10}}{10} \right]_0^{\pi} + \frac{1}{10} \int_0^{\pi} 5x^4 (\pi - x)^{10} \, dx \]

### Step 5: Evaluate the boundary terms again
Evaluating the boundary terms:
\[ \left[ -\frac{x^5 (\pi - x)^{10}}{10} \right]_0^{\pi} = 0 \]
Thus, we have:
\[ \int_0^{\pi} x^5 (\pi - x)^9 \, dx = \frac{1}{10} \int_0^{\pi} 5x^4 (\pi - x)^{10} \, dx = \frac{1}{2} \int_0^{\pi} x^4 (\pi - x)^{10} \, dx \]

### Step 6: Continue this process
Continuing this process, we will eventually find that:
\[ I = \frac{6!}{9 \cdot 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14} \int_0^{\pi} (\pi - x)^{14} \, dx \]

### Step 7: Evaluate the final integral
The integral \( \int_0^{\pi} (\pi - x)^{14} \, dx \) can be evaluated as:
\[ \int_0^{\pi} (\pi - x)^{14} \, dx = \int_0^{\pi} u^{14} \, du = \frac{\pi^{15}}{15} \]

### Step 8: Substitute back to find \( I \)
Substituting this back into our expression for \( I \):
\[ I = \frac{6!}{9 \cdot 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14} \cdot \frac{\pi^{15}}{15} \]

### Step 9: Calculate \( \frac{\pi^{15}}{\binom{15}{9} I} \)
Now, we need to compute:
\[ \frac{\pi^{15}}{\binom{15}{9} I} = \frac{\pi^{15}}{\frac{6!}{9 \cdot 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14} \cdot \frac{\pi^{15}}{15}} = \frac{15 \cdot 9 \cdot 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}{6! \cdot \binom{15}{9}} \]

### Step 10: Simplify the expression
Using \( \binom{15}{9} = \frac{15!}{9! \cdot 6!} \):
\[ \frac{\pi^{15}}{\binom{15}{9} I} = 9 \]

### Final Answer
Thus, the required answer is:
\[ \boxed{9} \]