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_(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 , prove that I_n+I_(n-2)=1/(n-1)dot

The value of int_(0)^(pi//4) (tan^(n)x+tan^(n-2)x)d(x-([x])/(1!)+([x]^(2))/(2!)-([x]^(3))/(3!)+....) where [x] is greatest function, is

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

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 (n^(n))/(n!)=(nx)/((n-1)!), x =

Find int tan^nx sec^2x\ dx, (n!=-1)

If A_n=int_0^(pi/2)(sin(2n-1)x)/(sinx)dx ,B_n=int_0^(pi/2)((sinn x)/(sinx))^2 dx for n inN , 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)

If S= tan^-1 (1/(n^2+n+1))+tan^-1 (1/(n^2+3n+3))+…+tan^-1 (1/(1+(n+19)(n+20))) then tan S is equal to (A) 20/(401+20n) (B) n/(n^2+20n+1) (C) n(401+20n) (D) 20/(n^2+n-1)

If n is not a multiple of 3, then the coefficient of x^n in the expansion of log_e (1+x+x^2) is : (A) 1/n (B) 2/n (C) -1/n (D) -2/n