If `I_n = int_0^(pi//4) tan^n xdx`, then

Options

A. `I_(10) + I_8 = 1/9`
B. `I_5 + I_7 = 1/6`
C. `I_12 + 2I_10 + I_8 = 20/99`
D. `I_6 - I_12 = 2/99`

Answer

To solve the integral \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \), we can use the technique of recurrence relations. Here’s a step-by-step solution:

### Step 1: Establish the recurrence relation
We can express \( I_n \) in terms of \( I_{n-1} \) and \( I_{n+1} \). We know that:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]
Using integration by parts or properties of integrals, we can derive:
\[ I_{n+1} + I_{n-1} = \frac{1}{n} \quad \text{for } n \geq 1 \]

### Step 2: Set up the equations
From the recurrence relation, we can express:
\[ I_{n+1} = \frac{1}{n} - I_{n-1} \]

### Step 3: Calculate specific values
We can calculate specific values of \( I_n \) for small integers to find a pattern.

1. **Base case**:
- \( I_0 = \int_0^{\frac{\pi}{4}} 1 \, dx = \frac{\pi}{4} \)
- \( I_1 = \int_0^{\frac{\pi}{4}} \tan x \, dx = -\ln(\cos x) \bigg|_0^{\frac{\pi}{4}} = -\ln(0) + \ln(1) = \frac{\pi}{4} \)

2. **Using the recurrence relation**:
- For \( n = 1 \):
\[ I_2 + I_0 = \frac{1}{1} \implies I_2 + \frac{\pi}{4} = 1 \implies I_2 = 1 - \frac{\pi}{4} \]
- For \( n = 2 \):
\[ I_3 + I_1 = \frac{1}{2} \implies I_3 + \frac{\pi}{4} = \frac{1}{2} \implies I_3 = \frac{1}{2} - \frac{\pi}{4} \]

### Step 4: Generalize the results
Continuing this process, we can find expressions for \( I_n \) in terms of \( I_{n-2} \) and so forth.

### Step 5: Solve for specific \( n \)
To find \( I_n \) for larger \( n \), we can continue applying the recurrence relation until we reach a known value.

### Step 6: Analyze the options
Given the recurrence relation and the calculated values, we can analyze the options provided in the problem statement to determine which are correct.