If `I_(n)=int_(0)^(pi//4) tan^(n)theta d theta`, then `I_(8)+I_(6)` equals

Options

A. `(1)/(4)`
B. `(1)/(5)`
C. `(1)/(6)`
D. `(1)/(7)`

Explanation

To solve the problem, we need to find the sum \( I_8 + I_6 \) where \( I_n = \int_0^{\frac{\pi}{4}} \tan^n \theta \, d\theta \).

### Step-by-Step Solution:

1. **Define the Integrals**:
\[ I_8 = \int_0^{\frac{\pi}{4}} \tan^8 \theta \, d\theta \]
\[ I_6 = \int_0^{\frac{\pi}{4}} \tan^6 \theta \, d\theta \]

2. **Combine the Integrals**:
\[ I_8 + I_6 = \int_0^{\frac{\pi}{4}} \tan^8 \theta \, d\theta + \int_0^{\frac{\pi}{4}} \tan^6 \theta \, d\theta = \int_0^{\frac{\pi}{4}} \left( \tan^8 \theta + \tan^6 \theta \right) d\theta \]

3. **Factor Out Common Terms**:
\[ I_8 + I_6 = \int_0^{\frac{\pi}{4}} \tan^6 \theta \left( \tan^2 \theta + 1 \right) d\theta \]

4. **Use the Trigonometric Identity**:
We know that \( 1 + \tan^2 \theta = \sec^2 \theta \):
\[ I_8 + I_6 = \int_0^{\frac{\pi}{4}} \tan^6 \theta \sec^2 \theta \, d\theta \]

5. **Substitution**:
Let \( t = \tan \theta \). Then, \( dt = \sec^2 \theta \, d\theta \). The limits change as follows:
- When \( \theta = 0 \), \( t = \tan(0) = 0 \)
- When \( \theta = \frac{\pi}{4} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \)

Thus, we can rewrite the integral:
\[ I_8 + I_6 = \int_0^1 t^6 \, dt \]

6. **Evaluate the Integral**:
\[ \int t^6 \, dt = \frac{t^{7}}{7} \bigg|_0^1 = \frac{1^7}{7} - \frac{0^7}{7} = \frac{1}{7} \]

7. **Final Result**:
Therefore, we have:
\[ I_8 + I_6 = \frac{1}{7} \]

### Conclusion:
The value of \( I_8 + I_6 \) is \( \frac{1}{7} \).