If I n -2 n x dx- thenA- 41- I 2- I 4- 1- I 3- I 5- 1- I 4- I 6 are in G-P-B- 1- I 2- I 4- 1- I 3- I 5- 1- I 4- I 6 are in A-P-C- I 2- I 4- I 3- I 5- I 4- I 6 are in A-P-D- I 2- I 4- I 3- I 52- I 4- I 6 are in G-P-

I_n+2+I_n=∫_π4^π2^n x·cosec^2x dx= [ ( x)^n+1n+1 ]^π2_π4 ⇒ I_n+2+I_n=1n+1 ⇒ I_2+I_4=13,I_3+I_5=14,I_4+I_6=15 So, 1I_2+I_4,1I_3+I_5,1I_4+I_6 are in A.P. ...

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Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

If I n =∫ππ/2...

Question

If $I_{n} = \frac{π}{2} \int \frac{π}{4} {cot}^{n} x d x$ , then

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

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

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I_{2} + I_{4}, (I_{3} + I_{5})^{2}, I_{4} + I_{6} are in G.P.

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Solution

The correct option is B $\frac{1}{I_{2} + I_{4}}, \frac{1}{I_{3} + I_{5}}, \frac{1}{I_{4} + I_{6}} are in A.P.$
$I_{n + 2} + I_{n} = \frac{π}{2} \int \frac{π}{4} {cot}^{n} x \cdot {cosec}^{2} x d x = {[\frac{- (cot x)^{n + 1}}{n + 1}]}_{\frac{π}{4}}^{\frac{π}{2}}$
$\Rightarrow I_{n + 2} + I_{n} = \frac{1}{n + 1}$
$\Rightarrow I_{2} + I_{4} = \frac{1}{3}, I_{3} + I_{5} = \frac{1}{4}, I_{4} + I_{6} = \frac{1}{5}$
So, $\frac{1}{I_{2} + I_{4}}, \frac{1}{I_{3} + I_{5}}, \frac{1}{I_{4} + I_{6}} are in A.P.$

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Similar questions

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I_{n} = \int_{0}^{π / 4} {tan}^{n} x d x,

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If In=π2∫π4cotnx dx, then

The correct option is B 1I2+I4,1I3+I5,1I4+I6 are in A.P.In+2+In=π2∫π4cotnx⋅cosec2x dx=[−(cotx)n+1n+1]π2π4 ⇒In+2+In=1n+1 ⇒I2+I4=13,I3+I5=14,I4+I6=15 So, 1I2+I4,1I3+I5,1I4+I6 are in A.P.

If $I_{n} = \frac{π}{2} \int \frac{π}{4} {cot}^{n} x d x$ , then