The greatest value of the function `f(x)=(sin2x)/(sin(x+pi/4))` on the interval `(0,pi/2)i s` `1/(sqrt(2))` (b) `sqrt(2)` (c) 1 (d) `-sqrt(2)`

The area enclosed by the curve y=sin x+cos x and y=|cos x-sin x| over the interval [0,(pi)/(2)] is 4(sqrt(2)-2) (b) 2sqrt(2)(sqrt(2)-1)2(sqrt(2)+1)(d)2sqrt(2)(sqrt(2)+1)

The difference between the greatest and the least values of the function f(x)=sin2x-x on [-(pi)/(2),(pi)/(2)]

Find the solutions of the equation, (log)_(sqrt(2)sin"x")(1+cosx)=2, in the interval x [0,2pi]

If x in(0,1), then greatest root of the equation sin2 pi x=sqrt(2)cos pi x is

The value of int_(-pi)^( pi)(x^(2))/(1+sin x+sqrt(1+sin^(2)x))dx is

The value of sin sqrt(x^(2) - pi^(2)/36) lies in the interval

Domain of the function f(x)=sqrt(cos(sin x))+sin(x^(2)-1) is [-1,1][-2,2]c*[-pi,sqrt(2)]uu[sqrt(2),pi][-sqrt(2),-sqrt(2)]

(dy)/(dx)=(sqrt((1-sin2x)/(1+sin2x))) is equal to,0

Prove that :int_(0)^((pi)/(2))sqrt(1-sin2x)dx=2(sqrt(2)-1)