I_(n)=int(t^(n))/(1+t^(2))dt" then "I_(n+2)=

If I_(1)=int_(x)^(1)(1)/(1+t^(2))dt and I_(2)=int_(1)^(1//x)(1)/(1+t^(2))dt for xgt0 , then

If I_(1)=int_(x)^(1)(1)/(1+t^(2)) dt and I_(2)=int_(1)^(1//x)(1)/(1+t^(2)) dt "for" x gt0 then,

If I_(1)=int_(x)^(1)(1)/(1+t^(2))dt and I_(2)=int_(1)^((1)/(2))(1)/(1+t^(2))dt for x>=0 then (A) I_(1)=I_(2)(B)I_(1)>I_(2)(C)I_(1)

If I_(1) = int_(x)^(1) (1)/(1+t^(2))dt and I_(2)=int_(1)^(1/x) (1)/(1+t^(2))dt for x gt0 , then

If I_(n)=int(x^(n))/(1+x^(2))dx, where n in N , then : I_(n+2)+I_(n)=

If I_(1)=int_(1//e)^(tanx)(t)/(1+t^(2))dtandI_(2)=int_(1//e)^(cotx)(dt)/(t(1+t^(2))) then the values of I_(1)+I_(2) is

If I_(n)=int(x^(n)dx)/(sqrt(x^(2)+a)) then prove that I_(n)+(n-1)/(n)al_(n-2)=(1)/(n)x^(n-1)*sqrt(x^(2)+a)

If n(ne 1) in N and I_(n) = int tan^(n) x dx then I_(n)+I_(n-2) =

If I_(n) = int_0^(pi/2) sin^(n)x dx then I_(n)/(I_(n-2)) =

I=int sqrt(1-t^(2))dt