If I _ n = _ 4 ^ 2 ^ n x dx , then : - Vidyadip

MCQ

If $I _{ n }=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^{ n } x dx ,$ then :

A
$\frac{1}{ I _{2}+ I _{4}}, \frac{1}{ I _{3}+ I _{5}}, \frac{1}{ I _{4}+ I _{6}}$ are in $G.P.$
B
$I _{2}+ I _{4}, I _{3}+ I _{5}, I _{4}+ I _{6}$ are in $A.P.$
C
$I _{2}+ I _{4},\left( I _{3}+ I _{5}\right)^{2}, I _{4}+ I _{6}$ are in $G.P.$
✓
$\frac{1}{ I _{2}+ I _{4}}, \frac{1}{ I _{3}+ I _{5}}, \frac{1}{ I _{4}+ I _{6}}$ are in $A.P.$

✓

Answer

Correct option: D.

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

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

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

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

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

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

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

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

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

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