解：∫(0,π/2)cos^5xsin2xdx=∫(0,π/2)cos^5x2sinxcosxdx=2∫(0,π/2)cos^6xsinxdx=-2∫(0,π/2)cos^6xd(cosx)=-2×1/7cos^7x(0,π/2)=-2/7[cos^7（π/2）-cos^70)=-2/7(0-1)=2/7
