Let `a_(n)=int_(0)^((pi)/(2))(1-cos2nx)/(1-cos2x)dx`, then `a_(1),a_(2),a_(2),"………."` are in

If a_(n)=int_(0)^((pi)/(2))(sin^(2)nx)/(sin^(2)x)dx, then det[[a_(1),a_(51),a_(101)a_(2),a_(52),a_(102)a_(3),a_(53),a_(103)]]

If a _(n) = int_(0) ^( (pi)/(2)) (sin^(2) nx)/( sin x) dx then a _(2) -a _(1) , a _(3) - a_(2) , a_(4) -a _(3),....... are in

A_(n)=int_(0)^((pi)/(2))(sin(2n-1)x)/(sin x)dx;B_(n)=int_(0)^((pi)/(2))((sin nx)/(sin x))^(2)dx; for n in N then A_(n+1)=A_(n)( b) B_(n+1)=B_(n)A_(n+1)-A_(n)=B_(n+1)(d)B_(n+1)-B=A_(n+1)

Given that a_(n)=int_(0)^(pi) (sin^(2){(n+1)x})/(sin2x)dx What is a_(n-1)-a_(n-4) equal to ?

If A_(n)=int_(0)^((pi)/(2))(sin(2n-1)x)/(sin x)dx,B_(n)=int_(0)^((pi)/(2))((sin nx)/(sin x))^(2)dx for n in N, 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)