If `2 sin^(2) ((pi//2) cos^(2) x)=1-cos (pi sin 2x), x ne (2n + 1) pi//2, n in I`, then `cos 2x` is equal to

Solve cos 2x=|sin x|, x in (-pi/2, pi) .

Solve cos 2x=|sin x|, x in (-pi/2, pi) .

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

sin (n +1) x sin (n +2) x + cos (n +1) x cos (n +2) x = cos x

Solve cos^(-1) (cos x) gt sin^(-1) (sin x), x in [0, 2pi]

If the area enclosed between f(x)=min (cos ^-1(cos x).cot ^-1(cot x)) and x -axis in x in(pi, 2 pi) is (pi^2)/k where k in N , then k is equal to

Evaluate: int_(-pi/2)^(pi/2)sqrt(cos^(2n-1)x-cos^(2n+1)x)xdx ,w h e r en in N

The value of int_(-(pi)/(2)) ^((pi)/(2)) sin^(2) x cos x dx is