Let `I_n = int_0^(pi//4) tan^n x dx (n gt 0 and in N)`, then

Options

A. `I_n = I_(n-2)`
B. `I_n + I_(n-2) = 1/(n -1)`
C. `I_n - I_(n - 2) = 1/(n -1)`
D. none of these

Answer

To solve the integral \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \) for \( n > 0 \) and \( n \in \mathbb{N} \), we can use a recursive relationship between \( I_n \) and \( I_{n-2} \).

### Step-by-Step Solution:

1. **Establish the Recursive Relationship:**
We start by noting that:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]
We can express \( \tan^n x \) as \( \tan^{n-2} x \cdot \tan^2 x \):
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \tan^2 x \, dx \]

2. **Use the Identity for \( \tan^2 x \):**
We know that \( \tan^2 x = \sec^2 x - 1 \). Thus:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x (\sec^2 x - 1) \, dx \]
This expands to:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx \]

3. **Evaluate the First Integral:**
The first integral can be simplified using the substitution \( u = \tan x \), which gives \( du = \sec^2 x \, dx \). The limits change from \( x = 0 \) to \( x = \frac{\pi}{4} \) (i.e., \( u = 0 \) to \( u = 1 \)):
\[ I_n = \int_0^1 u^{n-2} \, du - I_{n-2} \]

4. **Calculate the Integral:**
The integral \( \int_0^1 u^{n-2} \, du \) evaluates to:
\[ \int_0^1 u^{n-2} \, du = \frac{1}{n-1} \]
Therefore, we have:
\[ I_n = \frac{1}{n-1} - I_{n-2} \]

5. **Final Recursive Relation:**
Rearranging gives us the recursive relationship:
\[ I_n + I_{n-2} = \frac{1}{n-1} \]