यदि `I_(n) =underset(0)overset(pi//4)int tna ^(n) x dx, n in N,` तो

If I_(n)underset(0)overset(pi//4)inttan^(n) x dx, where n is a positive interger, then I_(10)+I_(8) is

If I_(n)=overset(pi//4)underset(0)int tan ^(n) x dx, lim_(n to oo) n(I_(n+1)+I_(n-1)) equals

I_(n) = int_0^(pi/4) tan^(n) x dx, n in N , then I_(10)+I_(8) =

Let I_(n)=int_(0)^(pi//4)tan^(n)xdx,n in N , Then

If I_(n)=int_(0)^(pi//4) tan^(n)x dx, (ngt1 is an integer ), then

Let I_(n)=int_(0)^(pi//2) sin^(n)x dx, nin N . Then

Let I_(n)=int_(0)^(pi//2) sin^(n)x dx, nin N . Then

Let I_(n) = overset(pi//4)underset(0)int tan^(n) x dx." Then "I_(2)+I_(4),I_(3)+I_(5),I_(4)+I_(6),I_(5)+I_(7),…. are in

If I_(n)=int_(0)^(pi//4)tan^(n)x dx, then 7(I_(6)+I_(8))=