Let `I_(n) = overset(pi//4)underset(0)int tan^(n) x dx." Then "I_(2)+I_(4),I_(3)+I_(5),I_(4)+I_(6),I_(5)+I_(7),….` are in

Options

A. A.P
B. G.P
C. H.P
D. none of these

Answer

To solve the problem, we need to analyze the integral \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \) and find the relationships between the sums \( I_2 + I_4, I_3 + I_5, I_4 + I_6, I_5 + I_7, \ldots \).

### Step-by-Step Solution:

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

2. **Calculate \( I_n + I_{n+2} \)**:
We can express \( I_n + I_{n+2} \) as follows:
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^n x \, dx + \int_0^{\frac{\pi}{4}} \tan^{n+2} x \, dx \]
This can be combined into a single integral:
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^n x (1 + \tan^2 x) \, dx \]

3. **Use the Identity**:
We know that \( 1 + \tan^2 x = \sec^2 x \). Thus, we can rewrite the integral:
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^n x \sec^2 x \, dx \]

4. **Substitution**:
Let \( u = \tan x \), then \( du = \sec^2 x \, dx \). The limits change as follows:
- When \( x = 0 \), \( u = 0 \)
- When \( x = \frac{\pi}{4} \), \( u = 1 \)

Therefore, we have:
\[ I_n + I_{n+2} = \int_0^1 u^n \, du = \left[ \frac{u^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1} \]

5. **Result for \( I_n + I_{n+2} \)**:
Thus, we find:
\[ I_n + I_{n+2} = \frac{1}{n+1} \]

6. **Calculate Specific Sums**:
Now, we can calculate:
- \( I_2 + I_4 = \frac{1}{2 + 1} = \frac{1}{3} \)
- \( I_3 + I_5 = \frac{1}{3 + 1} = \frac{1}{4} \)
- \( I_4 + I_6 = \frac{1}{4 + 1} = \frac{1}{5} \)
- \( I_5 + I_7 = \frac{1}{5 + 1} = \frac{1}{6} \)

7. **Identify the Series**:
The values we calculated are:
- \( I_2 + I_4 = \frac{1}{3} \)
- \( I_3 + I_5 = \frac{1}{4} \)
- \( I_4 + I_6 = \frac{1}{5} \)
- \( I_5 + I_7 = \frac{1}{6} \)

The reciprocals of these sums are:
- \( 3, 4, 5, 6 \)

Since \( 3, 4, 5, 6 \) are in Arithmetic Progression (AP), we conclude that:
\[ I_2 + I_4, I_3 + I_5, I_4 + I_6, I_5 + I_7 \text{ are in Harmonic Progression (HP)}. \]