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

Options

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

Explanation

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

### Step 1: Rewrite the Integral
We can rewrite the integral as follows:
\[ I = \int_{0}^{\frac{\pi}{2n}} \frac{1}{1 + \cos(nx)} \, dx \]

### Step 2: Use the Identity for Cosine
Using the identity \( \cos(nx) = \frac{\sin^2(nx) + \cos^2(nx)}{1} \), we can express the integral in terms of sine and cosine. However, a more useful transformation is to express it in terms of cotangent:
\[ I = \int_{0}^{\frac{\pi}{2n}} \frac{1}{1 + \cot(nx)} \, dx \]

### Step 3: Simplify the Denominator
We can rewrite \( \cot(nx) \) as \( \frac{\cos(nx)}{\sin(nx)} \):
\[ I = \int_{0}^{\frac{\pi}{2n}} \frac{\sin(nx)}{\sin(nx) + \cos(nx)} \, dx \]

### Step 4: Change of Variable
Now, we will perform a change of variable. Let \( t = nx \), then \( dx = \frac{dt}{n} \). The limits change as follows:
- When \( x = 0 \), \( t = 0 \)
- When \( x = \frac{\pi}{2n} \), \( t = \frac{\pi}{2} \)

Thus, the integral becomes:
\[ I = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} \frac{\sin(t)}{\sin(t) + \cos(t)} \, dt \]

### Step 5: Use Symmetry Property of Integrals
Using the property of integrals, we know:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
For our case, we have:
\[ I = \frac{1}{n} \int_{0}^{\frac{\pi}{2}} \frac{\cos(t)}{\sin(t) + \cos(t)} \, dt \]

### Step 6: Combine the Two Integrals
Let’s denote the two integrals:
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \frac{\sin(t)}{\sin(t) + \cos(t)} \, dt \]
\[ I_2 = \int_{0}^{\frac{\pi}{2}} \frac{\cos(t)}{\sin(t) + \cos(t)} \, dt \]
Thus, we have:
\[ I_1 + I_2 = \int_{0}^{\frac{\pi}{2}} 1 \, dt = \frac{\pi}{2} \]

### Step 7: Solve for \( I \)
Since \( I_1 = I \) and \( I_2 = I \), we have:
\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4} \]

### Step 8: Final Result
Now, substituting back, we find:
\[ I = \frac{\pi}{4n} \]

Thus, the value of the integral \( \int_{0}^{\frac{\pi}{2n}} \frac{1}{1 + \cos(nx)} \, dx \) is:
\[ \boxed{\frac{\pi}{4n}} \]