Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 31

int_0^(pi/2) (|sin(x)-cos(2x)|)dx
Since sin(x)-cos(2x)<=0,[0,pi/6]
and sin(x)-cos(2x)>=0,[pi/6,pi/2]
So, the integral can be split as,
=int_0^(pi/6)(-(sin(x)-cos(2x)))dx+int_(pi/6)^(pi/2)(sin(x)-cos(2x))dx
=int_0^(pi/6)(cos(2x)-sin(x))dx+int_(pi/6)^(pi/2)(sin(x)-cos(2x))dx
=[1/2sin(2x)+cos(x)]_0^(pi/6)+[-cos(x)-1/2sin(2x)]_(pi/6)^(pi/2)
=(1/2sin(pi/3)+cos(pi/6)-(1/2sin(0)+cos(0))+(-cos(pi/2)-1/2sin(pi))-(-cos(pi/6)-1/2sin(pi/3))
=(1/2*sqrt(3)/2+sqrt(3)/2)-(1)+(0)-(-sqrt(3)/2-1/2*sqrt(3)/2)
=(sqrt(3)/4+sqrt(3)/2-1+sqrt(3)/2+sqrt(3)/4)
=(3/2sqrt(3)-1)
~~1.598
Graph is attached. Integral is the sum of the region from (0 to pi/6) and (pi/6 to pi/2).