Answer
\[ = \frac{1}{{2\sqrt 2 }}\]
Work Step by Step
\[\begin{gathered}
\int_0^{\frac{3}{{\,\pi }}} {\sin \,\,\,\,\left( {2x - \frac{\pi }{4}} \right)\,\,\,dx} \hfill \\
\hfill \\
substituting\,\,\, \hfill \\
\hfill \\
\,\left( {2x - \frac{\pi }{4}} \right) = u \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
2dx = du\,\,or\,\,dx = \frac{{du}}{2}{\text{ }}limits\,\,also\,\,\,change\,\,as \hfill \\
\hfill \\
lower\,\,li\,mit\,\, = \,2\,\left( 0 \right) - \frac{\pi }{4} = - \frac{\pi }{4} \hfill \\
\hfill \\
upper\,\,\,li\,mit\, = \,2\,\left( {\frac{{3\pi }}{8}} \right) - \frac{\pi }{4} = \frac{\pi }{2} \hfill \\
\hfill \\
= \frac{1}{2}\int_{ - \pi /4}^{\pi /2} {\,\sin \,u\,du} \hfill \\
\hfill \\
integrate\,\,\,and\,evalute\,the{\text{ limits}} \hfill \\
\hfill \\
= \frac{1}{2}\left[ { - \cos u} \right]_{ - \pi /4}^{\pi /2} \hfill \\
\hfill \\
= - \frac{{\,\,\left[ {\cos \,u} \right]_{ - \frac{4}{\pi }}^{\frac{\pi }{2}}}}{2} = \frac{{\,\left( {0 - \frac{1}{{\sqrt 2 }}} \right)}}{2} \hfill \\
\hfill \\
= \frac{1}{{2\sqrt 2 }} \hfill \\
\end{gathered} \]
