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If An=int0^(pi/2)(sin(2n-1)x)/(sinx)dx ,...

If `A_n=int_0^(pi/2)(sin(2n-1)x)/(sinx)dx ,B_n=int_0^(pi/2)((sinn x)/(sinx))^2 dx` for `n inN `, Then
(A) `A_(n+1)=A_n` (B) `B_(n+1)=B_n` (C) `A_(n+1)-A_n=B_(n+1)` (D) `B_(n+1)-B_n=A_(n+1)`

A

`A_(n+1)=A_(n)`

B

`B_(n+1)=B_(n)`

C

`A_(n+1)-A_(n)=B_(n+1)`

D

`B_(n+1)-B_(n)=A_(n+1)`

Text Solution

Verified by Experts

The correct Answer is:
A, D

`A_(n+1)-A_(n)=int_(0)^(pi//2)(sin(2n+1)xsin(2n-1)x)/(sinx) dx`
`=int_(0)^(pi//2)2 cos 2nx dx=0`
or `A_(n+1)=A_(n)`
`B_(n+1)-B_(n)=int_(0)^(pi//2)(sin^(2)(n+1)x-sin^(2)nx)/(sin^(2)x)dx`
`=int_(0)^(pi//2) (sin(2n+1)x)/(sinx)dx=A_(n+1)`
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