solve that
Question:

If $I_{n}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^{n} x d x$, then :

1. $\frac{1}{\mathrm{I}_{2}+\mathrm{I}_{4}}, \frac{1}{\mathrm{I}_{3}+\mathrm{I}_{5}}, \frac{1}{\mathrm{I}_{4}+\mathrm{I}_{6}}$ are in G.P.

2. $\mathrm{I}_{2}+\mathrm{I}_{4}, \mathrm{I}_{3}+\mathrm{I}_{5}, \mathrm{I}_{4}+\mathrm{I}_{6}$ are in A.P.

3. $\mathrm{I}_{2}+\mathrm{I}_{4},\left(\mathrm{I}_{3}+\mathrm{I}_{5}\right)^{2}, \mathrm{I}_{4}+\mathrm{I}_{6}$ are in G.P.

4. $\frac{1}{I_{2}+I_{4}}, \frac{1}{I_{3}+I_{5}}, \frac{1}{I_{4}+I_{6}}$ are in A.P.

Correct Option: , 4

Solution:

$I_{n}=\int_{\pi / 4}^{\pi / 2} \cot ^{n} x d x=\int_{\pi / 4}^{\pi / 2} \cot ^{n-2} x\left(\operatorname{cosec}^{2} x-1\right) d x$

$\left.=-\frac{\cot ^{n-1} x}{n-1}\right]_{\pi / 4}^{\pi / 2}-I_{n-2}$

$=\frac{1}{n-1}-I_{n-2}$

$\Rightarrow \mathrm{I}_{\mathrm{n}}+\mathrm{I}_{\mathrm{n}-2}=\frac{1}{\mathrm{n}-1}$

$\Rightarrow \mathrm{I}_{2}+\mathrm{I}_{4}=\frac{1}{3}$

$\mathrm{I}_{3}+\mathrm{I}_{5}=\frac{1}{4}$

$\mathrm{I}_{4}+\mathrm{I}_{6}=\frac{1}{5}$

$\therefore \frac{1}{\mathrm{I}_{2}+\mathrm{I}_{4}}, \frac{1}{\mathrm{I}_{3}+\mathrm{I}_{5}}, \frac{1}{\mathrm{I}_{4}+\mathrm{I}_{6}}$ are in A.P.