Let f(x)=2cos2xsin2x−sinxsin2x2sin2xcosxsinx−cosx0. Then the value of ∫0π/2 f(x)+f'(x)dx is
π
π/2
2π
3π/2
Applying C1→C1−2sinxC3 and C2→C2+2cosxC3, we get
f(x)=20−sinx02cosxsinx−cosx0=2cos2x+2sin2x=2
∴ f'(x)=0
∴ ∫0π/2 f(x)+f'(x)=dx=∫0π/2 2dx=π