If `I_n=int_(pi/4)^(pi/2) (tanx)^-n dx(ngt1)`, then `I_n+I_(n+2)=` (A) `1/(n-1)` (B) `1/(n+1)` (C) `-1/(n+1)` (D) `1/n-1`

If I_(n)=int_(0)^(pi//4) tan^(n)x dx, (ngt1 is an integer ), then (a) I_(n)+I_(n-2)=1/(n+1) (b) I_(n)+I_(n-2)=1/(n-1) (c) I_(2)+I_(4),I_(6),……. are in H.P. (d) 1/(2(n+1))ltI_(n)lt1/(2(n-1))

If I_n=int( lnx)^n dx then I_n+nI_(n-1)

If n!=1, int_0^(pi/4) (tan^nx+tan^(n-2)x)d(x-[x])= (A) 1/(n-1) (B) 1/(n+1) (C) 1/n (D) 2/(n-1)

If I_n=int_0^(pi//4)tan^("n")x dx , prove that I_n+I_(n-2)=1/(n-1)dot

If I_(n)=int_(0)^(pi) e^(x)(sinx)^(n)dx , then (I_(3))/(I_(1)) is equal to

If = int_(0)^(1) x^(n)e^(-x)dx "for" n in N "then" I_(n)-nI_(n-1)=

Let I_(n) = int_(0)^(1)(1-x^(3))^(n)dx, (nin N) then

If = int_(1)^(e) (logx)^(n) dx, "then"I_(n)+nI_(n-1) is equal to

I_n = int_0^(pi/4) tan^n x dx , then the value of n(l_(n-1) + I_(n+1)) is

Let I_(n)=int_(0)^(pi//2)(sinx+cosx)^(n)dx(nge2) . Then the value of n. I_(n)-2(n-1)I_(n-1) is