Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 23

y=cos(x) , y=sin(2x) , x=0 , x=pi/2
Refer the attached image, y=cos(x) is plotted in red color and y=sin(2x) is plotted in blue color.
From graph,
cos(x) is above sin(2x) from 0 to pi/6
sin(2x) is above cos(x) fron pi6 to pi/2
Area of the region enclosed by the given curves A=int_0^(pi/6)(cos(x)-sin(2x))dx+int_(pi/6)^(pi/2)((sin(2x)-cos(x))dx
A=[sin(x)-(-cos(2x)/2)]_0^(pi/6)+[-cos(2x)/2-sin(x)]_(pi/6)^(pi/2)
A=[sin(x)+cos(2x)/2]_0^(pi/6)+[-cos(2x)/2-sin(x)]_(pi/6)^(pi/2)
A=(sin(pi/6)+cos(pi/3)/2)-(sin(0)+cos(0)/2)+(-cos(pi)/2-sin(pi/2))-(-cos(pi/3)/2-sin(pi/6))
A=(1/2+1/4)-(0+1/2)+(1/2-1)-(-1/4-1/2)
A=1/4-1/2+1/4+1/2
A=1/2