`I= int_(pi/2)^(pi/2) max (sin^-1 (sin x), f (x)), dx` where `f(x)={1, x>0 and -1, x<0` then I equal to

int_(0)^( pi)(x sin x)/(1+sin x)dx

int_(0)^(pi)(x sin x)/(1-sin x)dx=

int_((pi)/(5))^((3pi)/(10))(f(sin x))/(f(sin x) + f(cos x)) dx

int_(0)^( pi)(x sin(x))/(1+sin(x))dx

int_0^pi (x sin x)/(1+sin x) dx=pi/2 (pi-2)

I = int_ (1) ^ (2) | x sin pi x | dx

int_(0)^(pi//2) (sin x cos x) / ((1 + 2 sin x ) ( 1+ sin x))dx =

Find the value of int_(-(pi)/(2))^((pi)/(2)) f(x) dx, where f(x)={{:(sin^(2)x,"when "xgt0),(1-cosx,"when "xle0):}

If int_(0)^(pi)x f(sin x) dx = a int_(0)^(pi)f (sin x) dx , then a =

If int_(0)^(pi/2) f ( sin2 x ) sin x dx = A int_(0)^(pi/4) f ( cos 2 x ) cos x dx then the value of A is ( sqrt2 = 1.41)