Calc Integral: |sinx-cos2x| from 0 to \pi/2

[semc]

\int^{\pi/2}_{0}|sinx-cos2x| dx
Hi guys i can't seem to get the correct answer for this. What i did was draw the graph of sinx and cos2x and for 0 to \pi/6 i use cos2x-sinx since cos2x is higher. From \pi/6 to \pi/2 i use sinx-cos2x. However i am not able to get the same answer as my book can someone tell me is this the way to do integrals involving the absolute value?

 

[Dick]

You've split the integral up correctly. Your problem must be in evaluating the two integrals.