The curve `y=-x^2/2 + x + sin^2 x + sin^2((5pi)/2-x)` is symmetric with respect to line having equation :

The curve y=(x^(2))/(2)+x+1 is symmetric with respect to the line x-k, then k=

From the curve y=sin x, graph the functions. y= sin ((pi)/(2)-x) which is also cos x.

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

From the curve y=sin x, graph the functions. y= sin((pi)/(2)+x) which is cos x.

The subtangent at x = pi/2 on the curve y = x . sin x is

Area bounded by the curve y=sin^(2)x and lines x=(pi)/(2),x=pi and X-axis is

The area bounded by the curve y = sin^(2) x, 0 le x le pi/2 , X axis and the line x = pi/2 is

Let f (x) be a continous function such that the area bounded by the curve y =f (x) x-axis and the lines x=0 and x =a is (a ^(2))/( 2) + (a)/(2) sin a+ pi/2 cos a, then f ((pi)/(2)) is

The area bounded by the curve y=cosx and y=sin 2x, AA x in [(pi)/(6), (pi)/(2)] is equal to