Answer
$$\frac{{\sqrt 3 - 2}}{2}$$
Work Step by Step
$$\eqalign{
& \int_{\pi /2}^{2\pi /3} {\cos x} dx \cr
& {\text{integrate by using the basic integration rule }}\int {\cos x} dx = \sin x + C\,\,\,\left( {{\text{page 692}}} \right) \cr
& then \cr
& = \left( {\sin x} \right)_{\pi /2}^{2\pi /3} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \sin \left( {\frac{{2\pi }}{3}} \right) - \sin \left( {\frac{\pi }{2}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{\sqrt 3 }}{2} - 1 \cr
& = \frac{{\sqrt 3 - 2}}{2} \cr} $$
