2\sin x\cos x=(\sin x+\cos x)^2-1

Let t = \sin x+\cos x

\sin x\cos x = \dfrac{1}{2} t^2-\dfrac{1}{2}

y = \sin x+ \cos x+\sin x\cos x

= \dfrac{1}{2} t^2 +t-\dfrac{1}{2}

=\dfrac{1}{2}(t^2+2t+1)-1

t = \sin x+\cos x

=\sqrt{2} (\dfrac{\sqrt{2} }{2} \sin x+ \dfrac{\sqrt{2} }{2} \cos x)

= \sqrt{2} \sin (x+\dfrac{\pi}{4} )

Therefore, the domain of t in function (1) is ,

Evaluate the quadratic function (1), the function has the minimum value y=-1 when t = -1

The axis of symmetry of the parabola is x = -1

When t = \sqrt{2} , the function has its maximum value \sqrt{2} +1/2

The range of the function is [-1,\sqrt{2} +1/2 ]