`int_(0)^(pi//2n)(dx)/(1+(tan nx)^(n))` is equal to `n in N` :

Options

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

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2n}} \frac{dx}{1 + (\tan(nx))^n} \]

where \( n \in \mathbb{N} \), we will use a property of definite integrals and some trigonometric identities.

### Step 1: Use the property of definite integrals
We can use the property of definite integrals:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

In our case, \( a = 0 \) and \( b = \frac{\pi}{2n} \), so we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2n}} \frac{dx}{1 + (\tan(n(\frac{\pi}{2n} - x)))^n} \]

### Step 2: Simplify the expression inside the integral
Now, we need to simplify \( \tan(n(\frac{\pi}{2n} - x)) \):

\[ \tan(n(\frac{\pi}{2n} - x)) = \tan\left(\frac{\pi}{2} - nx\right) = \cot(nx) \]

Thus, we have:

\[ I = \int_{0}^{\frac{\pi}{2n}} \frac{dx}{1 + (\cot(nx))^n} \]

### Step 3: Rewrite the integral in terms of sine and cosine
Recall that \( \cot(nx) = \frac{\cos(nx)}{\sin(nx)} \), so:

\[ (\cot(nx))^n = \frac{(\cos(nx))^n}{(\sin(nx))^n} \]

This means:

\[ I = \int_{0}^{\frac{\pi}{2n}} \frac{dx}{1 + \frac{(\cos(nx))^n}{(\sin(nx))^n}} = \int_{0}^{\frac{\pi}{2n}} \frac{(\sin(nx))^n \, dx}{(\sin(nx))^n + (\cos(nx))^n} \]

### Step 4: Combine the two integrals
Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\frac{\pi}{2n}} \frac{dx}{1 + (\tan(nx))^n} \)
2. \( I = \int_{0}^{\frac{\pi}{2n}} \frac{(\sin(nx))^n \, dx}{(\sin(nx))^n + (\cos(nx))^n} \)

Adding these two expressions gives:

\[ 2I = \int_{0}^{\frac{\pi}{2n}} \left( \frac{dx}{1 + (\tan(nx))^n} + \frac{(\sin(nx))^n \, dx}{(\sin(nx))^n + (\cos(nx))^n} \right) \]

### Step 5: Evaluate the integral
Notice that the two terms in the integral combine nicely:

\[ 2I = \int_{0}^{\frac{\pi}{2n}} dx = \frac{\pi}{2n} \]

Thus, we find:

\[ I = \frac{\pi}{4n} \]

### Conclusion
The value of the integral is:

\[ I = \frac{\pi}{4n} \]