If `l_(n)=int_(0)^(pi//4) tan^(n)x dx, n in N "then" I_(n+2)+I_(n)` equals

Options

A. `(1)/(n)`
B. `(1)/(n-1)`
C. `(1)/(n+1)`
D. `(1)/(n+2)`

Explanation

To solve the problem, we need to find the expression for \( I_n + I_{n+2} \), where \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \).

### Step-by-Step Solution:

1. **Write the integral expression**:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]

2. **Express \( \tan^n x \) in terms of \( \tan^{n-2} x \)**:
We can rewrite \( \tan^n x \) as \( \tan^{n-2} x \cdot \tan^2 x \).
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \tan^2 x \, dx \]

3. **Substitute \( \tan^2 x \)**:
We know that \( \tan^2 x = \sec^2 x - 1 \), so we can substitute this into the integral:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot (\sec^2 x - 1) \, dx \]

4. **Split the integral**:
This gives us two integrals:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx \]
Let \( I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx \). Then we can write:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \sec^2 x \, dx - I_{n-2} \]

5. **Change of variable**:
For the first integral, we can use the substitution \( t = \tan x \), which gives \( dt = \sec^2 x \, dx \). The limits change from \( x = 0 \) to \( x = \frac{\pi}{4} \) resulting in \( t = 0 \) to \( t = 1 \):
\[ I_n = \int_0^1 t^{n-2} \, dt - I_{n-2} \]

6. **Evaluate the integral**:
The integral \( \int_0^1 t^{n-2} \, dt \) evaluates to:
\[ \int_0^1 t^{n-2} \, dt = \frac{1}{n-1} \quad \text{(for \( n > 1 \))} \]
Thus,
\[ I_n = \frac{1}{n-1} - I_{n-2} \]

7. **Express \( I_{n+2} \)**:
Similarly, we can express \( I_{n+2} \) using the same approach:
\[ I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^{n} x \cdot \sec^2 x \, dx - I_n \]
This leads to:
\[ I_{n+2} = \frac{1}{n+1} - I_n \]

8. **Combine the results**:
Now we can add \( I_n \) and \( I_{n+2} \):
\[ I_n + I_{n+2} = \left( \frac{1}{n-1} - I_{n-2} \right) + \left( \frac{1}{n+1} - I_n \right) \]
Rearranging gives:
\[ I_n + I_{n+2} = \frac{1}{n-1} + \frac{1}{n+1} - I_n - I_{n-2} \]

9. **Final expression**:
After simplification, we find that:
\[ I_n + I_{n+2} = \frac{1}{n+1} \]

### Conclusion:
Thus, the final result is:
\[ I_n + I_{n+2} = \frac{1}{n+1} \]