Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_{0.5}^{\left( {\pi + 1} \right)/2} {\sin \left( {2x - 1} \right)} dx \cr
& {\text{Let }}u = 2x - 1,{\text{ }}du = 2dx \cr
& {\text{The new limits of integration are:}} \cr
& x = \frac{{\pi + 1}}{2} \to u = 2\left( {\frac{{\pi + 1}}{2}} \right) - 1 = \pi \cr
& x = 0.5 \to u = 2\left( {0.5} \right) - 1 = 0 \cr
& {\text{Substituting}} \cr
& \int_{0.5}^{\left( {\pi + 1} \right)/2} {\sin \left( {2x - 1} \right)} dx = \int_0^\pi {\sin u} \left( {\frac{1}{2}} \right)du \cr
& = \frac{1}{2}\int_0^\pi {\sin u} du \cr
& {\text{Integrating}} \cr
& = - \frac{1}{2}\left[ {\cos u} \right]_0^\pi \cr
& = - \frac{1}{2}\left( {\cos \pi - \cos 0} \right) \cr
& = - \frac{1}{2}\left( { - 1 - 1} \right) \cr
& = 1 \cr} $$
