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)

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//2) sin^(n)x dx, nin N . Then

If A_n=int_0^(pi/2)(sin(2n-1)x)/(s in x)"dx";"B"_"n"=int_0^("pi"/2)(("s i n n x")/(s in x))^2"dx";"f o rn" in "N","t h e n" 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)

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

If I_(n) = int _(0)^(pi//4) tan^(n) theta " " d theta, "then "I_(8)+I_(6)=