If the slope of the curve `y = sin x + cos^2 x` is zero. Then find the value of `x` in `x in [0, pi/2]`

The slope of the curve y=sin x+cos^(2)x is zero at the point where

Let the slope of the curve y = cos^(-1) (sin x) be tan theta : Then the value of theta in the interval (0, pi) is

Find the slope of normal to the curve y = sin^(2) x at x = pi//4 .

The area of the region enclosed between the curves y= sinx + cos x and y= | cos x - sin x|, x in [ 0, pi/2] is ……..Sq. units.

Find the area enclosed between the curve y= sin x and y = cos x that lies between thhe lines x= 0 and x= pi/2 .

The area enclosed by the curves: y=sin x+ cos x and y=cos x- sin x| over the interval [0,pi/2] is:

The area enclosed by the curves y=sin x + cos x and y |cos x -sin x| over the interval [0,pi//2] is :

The slope of the tangent to the curve y = x + sin x cos x at x = (pi)/( 2) is

The slope of normal to the curve y= sin^(2)x and x= (pi)/(4) is

The area enclosed by the curves y=sin x+cos x and y=|cos x-sin x| over the interval [0,(pi)/(2)]