If $I_{n}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^{n} x d x$, then :
Correct Option: , 4
$I_{n}=\int_{\pi / 4}^{\pi / 2} \cot ^{n} x d x=\int_{\pi / 4}^{\pi / 2} \cot ^{n-2} x\left(\operatorname{cosec}^{2} x-1\right) d x$
$\left.=-\frac{\cot ^{n-1} x}{n-1}\right]_{\pi / 4}^{\pi / 2}-I_{n-2}$
$=\frac{1}{n-1}-I_{n-2}$
$\Rightarrow \mathrm{I}_{\mathrm{n}}+\mathrm{I}_{\mathrm{n}-2}=\frac{1}{\mathrm{n}-1}$
$\Rightarrow \mathrm{I}_{2}+\mathrm{I}_{4}=\frac{1}{3}$
$\mathrm{I}_{3}+\mathrm{I}_{5}=\frac{1}{4}$
$\mathrm{I}_{4}+\mathrm{I}_{6}=\frac{1}{5}$
$\therefore \frac{1}{\mathrm{I}_{2}+\mathrm{I}_{4}}, \frac{1}{\mathrm{I}_{3}+\mathrm{I}_{5}}, \frac{1}{\mathrm{I}_{4}+\mathrm{I}_{6}}$ are in A.P.
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