`I_(n)=int_(pi/4)^(pi/2)(cot^(n)x)dx` , then :

Options

A. `1/(I_(2)+I_(4)),1/(I_(3)+I_(5)),1/(I_(4)+I_(6))` are in G.P.
B. `I_(2) + I_(4) , I_(3) + I_(5) , I_(4) + I_(6) ` are in A.P.
C. `I_(2) + I_(4) , (I_(3) + I_(5))^2 , I_(4) + I_(6) ` are in G.P.
D. `1/(I_(2)+I_(4)),1/(I_(3)+I_(5)),1/(I_(4)+I_(6))` are in A.P.

Answer

To solve the integral \( I_n = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^n x \, dx \), we can use the properties of integrals and some relationships between different \( I_n \) values.

### Step 1: Establish the relationship between \( I_n \) and \( I_{n-2} \)

We can express \( I_n \) in terms of \( I_{n-2} \) using integration by parts or a reduction formula. We can write:

\[ I_n = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^n x \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \cdot \cot^2 x \, dx \]

### Step 2: Rewrite \( \cot^2 x \)

Recall that \( \cot^2 x = \csc^2 x - 1 \). Therefore, we can split the integral:

\[ I_n = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \cdot (\csc^2 x - 1) \, dx \]

This gives us:

\[ I_n = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \cdot \csc^2 x \, dx - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \, dx \]

### Step 3: Define the integrals

Let \( I_{n-2} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \, dx \) and \( I_{n} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^n x \, dx \).

Thus, we have:

\[ I_n = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \cdot \csc^2 x \, dx - I_{n-2} \]

### Step 4: Evaluate the first integral

Using the substitution \( t = \cot x \), we have \( dt = -\csc^2 x \, dx \). The limits change as follows:

- When \( x = \frac{\pi}{4} \), \( t = 1 \)
- When \( x = \frac{\pi}{2} \), \( t = 0 \)

Thus, we can rewrite the integral:

\[ I_n = \int_{1}^{0} t^{n-2} (-dt) - I_{n-2} = \int_{0}^{1} t^{n-2} \, dt - I_{n-2} \]

Calculating the integral:

\[ \int_{0}^{1} t^{n-2} \, dt = \frac{1}{n-1} \]

So we have:

\[ I_n = \frac{1}{n-1} - I_{n-2} \]

### Step 5: Establish a recurrence relation

Rearranging gives us:

\[ I_n + I_{n-2} = \frac{1}{n-1} \]

### Step 6: Use the recurrence relation

Using this recurrence relation, we can find specific values:

1. For \( n = 2 \):
\[ I_2 + I_0 = \frac{1}{1} \implies I_2 + I_0 = 1 \]

2. For \( n = 4 \):
\[ I_4 + I_2 = \frac{1}{3} \]

3. For \( n = 3 \):
\[ I_3 + I_1 = \frac{1}{2} \]

4. For \( n = 5 \):
\[ I_5 + I_3 = \frac{1}{4} \]

5. For \( n = 6 \):
\[ I_6 + I_4 = \frac{1}{5} \]

### Conclusion

From these equations, we can establish relationships between the integrals \( I_n \) and find specific values. The relationships show that \( I_2, I_4, I_3, I_5, I_6 \) are in arithmetic progression or geometric progression based on the derived equations.