Let `I_(n) = int _(0)^(pi) (sin (n + (1)/(2))x )/(sin ((x)/(2)))dx` where `n in N ` U ` {0}. `If ` l _(1)^(2)+l _(2)^(2) +l_(3)^(2)+ ……. + l _(20)^(2) =mpi^(2),` then find the largest prime factor of m.

Answer

To solve the given problem, we need to evaluate the integral \( I_n = \int_0^{\pi} \frac{\sin\left(n + \frac{1}{2}\right)x}{\sin\left(\frac{x}{2}\right)} \, dx \) for \( n \in \mathbb{N} \cup \{0\} \) and find the largest prime factor of \( m \) where \( l_1^2 + l_2^2 + l_3^2 + \ldots + l_{20}^2 = m\pi^2 \).

### Step 1: Establish the integral \( I_n \)

We start with the integral:
\[ I_n = \int_0^{\pi} \frac{\sin\left(n + \frac{1}{2}\right)x}{\sin\left(\frac{x}{2}\right)} \, dx \]

### Step 2: Use the property of the integral

We can use the property of the integral for \( I_{n+1} \):
\[ I_{n+1} = \int_0^{\pi} \frac{\sin\left(n + \frac{3}{2}\right)x}{\sin\left(\frac{x}{2}\right)} \, dx \]

### Step 3: Establish a relationship between \( I_n \) and \( I_{n+1} \)

Subtracting \( I_n \) from \( I_{n+1} \):
\[ I_{n+1} - I_n = \int_0^{\pi} \left( \frac{\sin\left(n + \frac{3}{2}\right)x - \sin\left(n + \frac{1}{2}\right)x}{\sin\left(\frac{x}{2}\right)} \right) dx \]

Using the sine subtraction formula:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
where \( A = \left(n + \frac{3}{2}\right)x \) and \( B = \left(n + \frac{1}{2}\right)x \).

### Step 4: Simplify the expression

This gives:
\[ I_{n+1} - I_n = 2 \int_0^{\pi} \frac{\cos\left(n + 1\right)x \sin\left(x\right)}{\sin\left(\frac{x}{2}\right)} \, dx \]

### Step 5: Evaluate \( I_0 \)

Now, let's evaluate \( I_0 \):
\[ I_0 = \int_0^{\pi} \frac{\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \, dx = \int_0^{\pi} 1 \, dx = \pi \]

### Step 6: Find \( I_n \) for \( n \geq 1 \)

Since all \( I_n \) are equal for \( n = 0, 1, 2, \ldots, 20 \):
\[ I_n = \pi \quad \text{for all } n \]

### Step 7: Calculate \( l_1^2 + l_2^2 + \ldots + l_{20}^2 \)

Since \( I_n = \pi \):
\[ l_1^2 + l_2^2 + \ldots + l_{20}^2 = 20 \cdot \pi^2 \]
Thus, we have:
\[ m \pi^2 = 20 \pi^2 \implies m = 20 \]

### Step 8: Find the largest prime factor of \( m \)

The prime factorization of \( 20 \) is:
\[ 20 = 2^2 \times 5 \]
The largest prime factor of \( 20 \) is \( 5 \).

### Final Answer

The largest prime factor of \( m \) is:
\[ \boxed{5} \]