`int_0^x|sint|dt`, where `x in (2npi,(2n+1)pi)`,`ninN`,is equal to (A) `4n-cosx` (B) `4n-sinx` (C) `4 n+1-cosx` (D) `4n-1-cosx`

int_(0)^(x)[sin t]dt, where x in(2n pi,(2n+1)pi),n in N, and [.] denotes the greatest integer function is equal to -n pi(b)-(n+1)pi2n pi(d)-(2n+1)pi

Evaluate: int_(0)^(x)[cos t]dt where x in(2n pi,4n+1(pi)/(2)),n in N, and.] denotes the greatest integer function.

The value of int_0^x[cost]dt ,x in [(4n+1)pi/2,(4n+3)pi/2]a n dn in N , is equal to where [.] represents greatest integer function. pi/2(2n-1)-2x pi/2(2n-1)+x pi/2(2n+1)-x (d) pi/2(2n+1)+x

"If" I=int_0^x(t^2dt)/((x-sinx)sqrt(a+t))a n d(lim)_(xvec0)I=1, then a is equal to : 1 (2) 2 (3) 4 (4) 0 (5) 3

Solve : 4sinx cosx +2sinx+2cosx+1=0

if f(x)=|[cosx,1,0],[1,2cosx,1],[0,1,2cosx]| then int_0^(pi/2) f(x)dx is equal to (A) 1/4 (B) -1/3 (C) 1/2 (D) 1

If I_n=int_0^(pi/4) tan^nx then lim_(nrarroo)n(I_n+I_(n-2)) equals (A) 1/2 (B) 1 (C) oo (D) 0

I_n=int_0^(pi//4) tan^n x dx, then lim_(ntooo) n [I_n + I_(n+2)] is equal to (i)1/2 (ii)1 (iii)infty (iv) 0

int_0^(2npi) {|sinx|-|1/2sinx|}dx= (A) n (B) 2n (C) -2n (D) 1/2