If `I _(n)=int _(0)^(pi) (sin (2nx))/(sin 2x)dx, ` then the value of `I _( n +(1)/(2))` is equal to `(n in I)`:

Options

A. 1: `(npi)/(2()`
B. 2: `pi`
C. 3: `pi/2`
D. 4: 0

Answer

To solve the problem, we need to evaluate the integral

\[ I_n = \int_0^{\pi} \frac{\sin(2nx)}{\sin(2x)} \, dx \]

and specifically find the value of \(I_{n + \frac{1}{2}}\).

### Step 1: Substitute \(n\) with \(n + \frac{1}{2}\)

We start by substituting \(n\) with \(n + \frac{1}{2}\) in the integral:

\[ I_{n + \frac{1}{2}} = \int_0^{\pi} \frac{\sin(2(n + \frac{1}{2})x)}{\sin(2x)} \, dx \]

### Step 2: Simplify the sine function

Using the identity for sine, we can expand the sine term:

\[ \sin(2(n + \frac{1}{2})x) = \sin(2nx + x) = \sin(2nx)\cos(x) + \cos(2nx)\sin(x) \]

Thus, we can rewrite the integral as:

\[ I_{n + \frac{1}{2}} = \int_0^{\pi} \frac{\sin(2nx)\cos(x) + \cos(2nx)\sin(x)}{\sin(2x)} \, dx \]

### Step 3: Split the integral

We can split the integral into two parts:

\[ I_{n + \frac{1}{2}} = \int_0^{\pi} \frac{\sin(2nx)\cos(x)}{\sin(2x)} \, dx + \int_0^{\pi} \frac{\cos(2nx)\sin(x)}{\sin(2x)} \, dx \]

### Step 4: Simplify each integral

Using the identity \(\sin(2x) = 2\sin(x)\cos(x)\), we can rewrite the integrals:

1. For the first integral:

\[ \int_0^{\pi} \frac{\sin(2nx)\cos(x)}{\sin(2x)} \, dx = \int_0^{\pi} \frac{\sin(2nx)}{2\sin(x)} \, dx \]

2. For the second integral:

\[ \int_0^{\pi} \frac{\cos(2nx)\sin(x)}{\sin(2x)} \, dx = \int_0^{\pi} \frac{\cos(2nx)}{2\cos(x)} \, dx \]

### Step 5: Evaluate the integrals

Both integrals have a period of \(2\pi\). Therefore, over the interval from \(0\) to \(\pi\), they will cancel out due to symmetry properties of sine and cosine functions.

Thus, we conclude:

\[ I_{n + \frac{1}{2}} = 0 \]

### Final Result

The value of \(I_{n + \frac{1}{2}}\) is:

\[ \boxed{0} \]