Chapter 13 - Multiple Integration - 13.2 Double Integrals over General Regions - 13.2 Exercises - Page 981: 45

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$$\eqalign{ & \int_0^{\pi /2} {\int_y^{\pi /2} {6\sin \left( {2x - 3y} \right)dx} dy} \cr & {\text{Integrate with respect to }}x \cr & = \int_0^{\pi /2} {\left[ { - \frac{6}{2}\cos \left( {2x - 3y} \right)} \right]_y^{\pi /2}dy} \cr & {\text{Evaluating the limits}} \cr & = - \frac{6}{2}\int_0^{\pi /2} {\left[ {\cos \left( {2\left( {\frac{\pi }{2}} \right) - 3y} \right) - \cos \left( {2\left( y \right) - 3y} \right)} \right]dy} \cr & = - \frac{6}{2}\int_0^{\pi /2} {\left[ {\cos \left( {\pi - 3y} \right) - \cos \left( { - y} \right)} \right]dy} \cr & = - \frac{6}{2}\int_0^{\pi /2} {\left[ {\cos \left( {\pi - 3y} \right) - \cos y} \right]dy} \cr & {\text{Integrate}} \cr & = - \frac{6}{2}\left[ { - \frac{1}{3}\sin \left( {\pi - 3y} \right) - \sin y} \right]_0^{\pi /2} \cr & {\text{Evaluating}} \cr & = - \frac{6}{2}\left[ { - \frac{1}{3}\sin \left( {\pi - \frac{3\pi }{2}} \right) - \sin \frac{\pi }{2}} \right] + 3\left[ { - \frac{1}{3}\sin \left( {\pi - 3\left( 0 \right)} \right) - \sin 0} \right] \cr & = - \frac{6}{2}\left[ { - \frac{1}{3}\sin \left( {-\frac{\pi }{2}} \right) - \sin \frac{\pi }{2}} \right] + 3\left[ 0 \right] \cr & = - \frac{6}{2}\left[ { \frac{1}{3} - 1} \right] \cr & = 2 \cr} $$