Prove that `I_(1),I_(2),I_(3)"..."` form an AP, if
(i) `I_(n)=int_(0)^(pi)(sin2nx)/(sinx)dx`
(ii) `I_(n)=int_(0)^(pi)((sinnx)/(sinx))^(2)dx`.

Explanation

(i)`I_(n)=int_(0)^(pi)(sin2nx)/(sinx)dx=int_(0)^(pi)(sin2n(pi-x))/(sinx)dx`
`=int_(0)^(pi)(sin(2npi-2nx))/(sinx)dx`
`I_(n)=-int_(0)^(pi)(sin2nx)/(sinx)dx`
`I_(n)=-I_(n)implies2I_(n)=0 implies I_(n)=0`
`:.I_(1)=I_(2)=I_(3)="..."=0`
which is a constant series.
`:.` This series is AP with common difference 0 and first term o.
(ii) `I_(n)=int_(0)^(pi)(sin^(2)nx)/(sin^(2)x)dx`
Let `f(x)=int_(0)^(pi)(sin^(2)nx)/(sin^(2)x)`
Hence, `f(pi-x)=f(x)`
So, `I_(n)=2int_(0)^((pi)/(2))(sin^(2)nx)/(sin^(2)x)dx`
Now, `I_(n+1)+I_(n-1)-2I_(n)`
`=2int_(0)^((pi)/(2)){((sin^(2)(n+1)x-sin^(2)nx))/(sin^(2)x)+((sin^(2)(n-1)x-sin^(2)nx))/(sin^(2)x)}dx`
`=2int_(0)^((pi)/(2))(sin(2n+1)xsinx-sin(2n-1)xsin x)/(sin^(2)x)dx`
`=2int_(0)^((pi)/(2))(sin(2n+1)x-sin(2n-1)x)/(sinx)dx`
`=2int_(0)^((pi)/(2))(2cos 2nx sinx)/(sinx)dx`
`=4int_(0)^((pi)/(2))2cos 2nx dx=(4)/(2n)[sin2nx]_(0)^((pi)/(2))=(2)/(n)*0=0`
`:.I_(n+1)+I_(n-1)=2I_(n):.I_(1),I_(2),I_(3),"......"`.