Answer
$$\frac{{\sqrt 3 - \sqrt 2 }}{2}$$
Work Step by Step
$$\eqalign{
& \int_{\pi /6}^{\pi /4} {\sin x} dx \cr
& {\text{integrate by using the basic integration rule }}\int {\sin x} dx = - \cos x + C\,\,\,\left( {{\text{page 692}}} \right) \cr
& then \cr
& = - \left( {\cos x} \right)_{\pi /6}^{\pi /4} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = - \left( {\cos \frac{\pi }{4} - \cos \frac{\pi }{6}} \right) \cr
& {\text{simplifying}} \cr
& = \cos \frac{\pi }{6} - \cos \frac{\pi }{4} \cr
& = \frac{{\sqrt 3 }}{2} - \frac{{\sqrt 2 }}{2} \cr
& = \frac{{\sqrt 3 - \sqrt 2 }}{2} \cr} $$
