[" If "I_(n)=int_(0)^( pi/2)(sin(2n-1)x)/(sin x)dx],[" and "a_(n)=int_(0)^( pi/2)((sin n theta)/(sin theta))^(2)d theta," then "a_(n+1)-a_(n)=]

If I_n=int_0^(pi/2) (sin(2n-1)x)/sinx)dx , and a_n=int_0^(pi/2) ((sinntheta)/sintheta)^2 d theta , then a_(n+1)-a_n= (A) I_n (B) 2I_n (C) I_(n+1) (D) 0

If quad 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

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)

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)

If I_(n)=int_(0)^(2 pi)(cos(n theta))/(cos theta)d(theta), where n in N then.

The value of I(n)=int_(0)^( pi)(sin^(2)n theta)/(sin^(2)theta)d theta is (AA n in N)

Let I_(n) = int_(0)^(pi)(sin^(2)(nx))/(sin^(2)x)dx , n in N then

int_(-pi//2)^(pi//2)(sin^(2n-1)x)/(1+cos^(2n)x)dx=