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Question: I<sub>1(n)</sub> =\(\int _ { 0 } ^ { \pi / 2 } \frac { \sin ( 2 n - 1 ) } { \sin x }\)dx, I<sub>2(n)...

I1(n) =∫0π/2sin⁡(2n−1)sin⁡x\int _ { 0 } ^ { \pi / 2 } \frac { \sin ( 2 n - 1 ) } { \sin x }dx, I2(n) = dx, n ĪN, then-

A

I2(n +1) – I2(n) = I1(n)

B

I2(n +1) – I2(n) = I1(n + 1)

C

I2(n +1) + I1(n) = I2(n)

D

I2(n +1) + I1(n+1) = I2(n)

Answer

I2(n +1) – I2(n) = I1(n + 1)

Explanation

Solution

I2(n) – I2(n – 1) =

=

=

= I1(n)

Ž I2(n+1) – I2(n) = I1(n+1).

Hence (2) is the correct answer.