Let `f` be the function defined on `[-pi,pi]` given by `f(0)=9` and `f(x)=sin((9x)/2)/sin(x/2)` for `x!=0`. The value of `2/pi int_-pi^pif(x)dx` is (asked as Match the following question)

Options

A. 0
B. 2
C. 4
D. 6

Explanation

To solve the problem, we need to evaluate the integral \( I = \frac{2}{\pi} \int_{-\pi}^{\pi} f(x) \, dx \) where \( f(x) = \frac{\sin\left(\frac{9x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \) for \( x \neq 0 \) and \( f(0) = 9 \).

### Step-by-Step Solution:

1. **Recognize the symmetry of the function**:
Since \( f(x) \) is defined for \( x \neq 0 \) and \( f(0) = 9 \), we first check if \( f(x) \) is an even function. We calculate \( f(-x) \):
\[ f(-x) = \frac{\sin\left(-\frac{9x}{2}\right)}{\sin\left(-\frac{x}{2}\right)} = \frac{-\sin\left(\frac{9x}{2}\right)}{-\sin\left(\frac{x}{2}\right)} = \frac{\sin\left(\frac{9x}{2}\right)}{\sin\left(\frac{x}{2}\right)} = f(x) \]
Thus, \( f(x) \) is an even function.

2. **Use the property of definite integrals**:
Since \( f(x) \) is even, we can simplify the integral:
\[ \int_{-\pi}^{\pi} f(x) \, dx = 2 \int_{0}^{\pi} f(x) \, dx \]
Therefore, we can write:
\[ I = \frac{2}{\pi} \cdot 2 \int_{0}^{\pi} f(x) \, dx = \frac{4}{\pi} \int_{0}^{\pi} f(x) \, dx \]

3. **Change of variables**:
Let \( x = 2t \) (where \( dx = 2dt \)). The limits change from \( x = 0 \) to \( x = \pi \) to \( t = 0 \) to \( t = \frac{\pi}{2} \):
\[ I = \frac{4}{\pi} \int_{0}^{\frac{\pi}{2}} f(2t) \cdot 2 \, dt = \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} f(2t) \, dt \]
Now, substituting \( f(2t) \):
\[ f(2t) = \frac{\sin(9t)}{\sin(t)} \]
So we have:
\[ I = \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{\sin(9t)}{\sin(t)} \, dt \]

4. **Use the formula for integrals of this form**:
The integral \( \int_{0}^{\frac{\pi}{2}} \frac{\sin(nt)}{\sin(t)} \, dt \) is known to equal \( \frac{\pi}{2} \) for integer \( n \). Thus:
\[ \int_{0}^{\frac{\pi}{2}} \frac{\sin(9t)}{\sin(t)} \, dt = \frac{\pi}{2} \]

5. **Substituting back into the expression for \( I \)**:
\[ I = \frac{8}{\pi} \cdot \frac{\pi}{2} = 4 \]

6. **Final result**:
Therefore, the value of \( 2/\pi \int_{-\pi}^{\pi} f(x) \, dx \) is:
\[ \boxed{4} \]