## Calculus: Early Transcendentals (2nd Edition)

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