If `I_(n)=int_(0)^(pi)(1-sin2nx)/(1-cos2x)dx` then `I_(1),I_(2),I_(3),"….."` are in

Options

A. AP
B. GP
C. HP
D. None of these

Answer

A

Explanation

`:. I_(n)=int_(0)^(pi)(1-sin2nx)/(1-cos2x)dx " " implies I_(n)=int_(0)^(pi)(1-sin2nx)/(2sin^(2)x)dx`
`implies I_(n+1)+I_(n-1)-2 I_(n) [ 1-sin2 (n+1)x+1-sin 2(n-1)x-2]`
`=(1)/(2)int_(0)^(pi)(+sin2nx " "]/(sin^(2)x)dx`
`=(1)/(2)int_(0)^(pi)([sin2nx-sin 2 (n+1)x]+[sin2nx-sin 2 (n-1)x])/(sin^(2)x)dx`
`=(1)/(2)int_(0)^(pi)(-2cos(2n+1)xsin(x)+2cos(2n-1)xsinx)/(sin^(2)x)dx`
`=int_(0)^(pi)(sinx[cos(2n-1)x-cos(2n+1)x])/(sin^(2)x)dx`
`=int_(0)^(pi)(2sin2nxsinx)/(sinx)dx=2int_(0)^(pi)sin2nxdx=(2)/(2n)[-cos2nx]_(0)^(pi)`
`=-(1)/(n)(1-1)=0` `:. I_(n+1)+I_(n-1)=2I_(n)`
`implies I_(n-1)+I_(n),I_(n+1)` are in AP.
`:.I_(1),I_(2),I_(3)"...."` are in AP.