The value of `int_(0)^(pi//2n) (1)/(1+cot nx)dx`, is

Options

A. 0
B. `(pi)/(4n)`
C. `(pi)/(2n)`
D. `(pi)/(2)`

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2n}} \frac{1}{1 + \cot(nx)} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = nx \). Then, differentiating both sides gives us:
\[ dx = \frac{dt}{n} \]
Now, we need to change the limits of integration. When \( x = 0 \), \( t = 0 \), and when \( x = \frac{\pi}{2n} \), \( t = \frac{\pi}{2} \). Thus, the integral becomes:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \cot(t)} \cdot \frac{dt}{n} \]

### Step 2: Simplifying the Integral
Now, we can rewrite the integral:
\[ I = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \cot(t)} \, dt \]
Recall that \( \cot(t) = \frac{\cos(t)}{\sin(t)} \), so:
\[ 1 + \cot(t) = 1 + \frac{\cos(t)}{\sin(t)} = \frac{\sin(t) + \cos(t)}{\sin(t)} \]
Thus, we can rewrite the integral:
\[ I = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} \frac{\sin(t)}{\sin(t) + \cos(t)} \, dt \]

### Step 3: Using Symmetry
To evaluate this integral, we can use the property of integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, let’s apply this property:
\[ I = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} \frac{\sin\left(\frac{\pi}{2} - t\right)}{\sin\left(\frac{\pi}{2} - t\right) + \cos\left(\frac{\pi}{2} - t\right)} \, dt \]
This simplifies to:
\[ I = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} \frac{\cos(t)}{\cos(t) + \sin(t)} \, dt \]

### Step 4: Adding the Two Integrals
Now, we have two expressions for \( I \):
1. \( I = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} \frac{\sin(t)}{\sin(t) + \cos(t)} \, dt \)
2. \( I = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} \frac{\cos(t)}{\sin(t) + \cos(t)} \, dt \)

Adding these two integrals gives:
\[ 2I = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} \frac{\sin(t) + \cos(t)}{\sin(t) + \cos(t)} \, dt = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} 1 \, dt \]
This simplifies to:
\[ 2I = \frac{1}{n} \cdot \left[ t \right]_{0}^{\frac{\pi}{2}} = \frac{1}{n} \cdot \frac{\pi}{2} \]

### Step 5: Solve for \( I \)
Now, solving for \( I \):
\[ 2I = \frac{\pi}{2n} \implies I = \frac{\pi}{4n} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{4n}} \]