G
Guest
Guest
Please would you be kind enough to look at this for me? Thank You.
\(\displaystyle \L
\int_0^\pi {2\cos (3x + \frac{\pi }{2}} )dx\)
\(\displaystyle \L\
= {\textstyle{2 \over 3}}\left[ {\sin (3x + \frac{\pi }{2})} \right]_0^\pi = {\textstyle{2 \over 3}}\left[ {\sin (3(\pi ) + \frac{\pi }{2}) - \sin (3(0) + \frac{\pi }{2})} \right]\)
\(\displaystyle \L\
= {\textstyle{2 \over 3}}\left[ {\sin \frac{{7\pi }}{2} - \sin \frac{\pi }{2}} \right] = {\textstyle{2 \over 3}}\sin 3\pi\)
:?:
Thank You.
\(\displaystyle \L
\int_0^\pi {2\cos (3x + \frac{\pi }{2}} )dx\)
\(\displaystyle \L\
= {\textstyle{2 \over 3}}\left[ {\sin (3x + \frac{\pi }{2})} \right]_0^\pi = {\textstyle{2 \over 3}}\left[ {\sin (3(\pi ) + \frac{\pi }{2}) - \sin (3(0) + \frac{\pi }{2})} \right]\)
\(\displaystyle \L\
= {\textstyle{2 \over 3}}\left[ {\sin \frac{{7\pi }}{2} - \sin \frac{\pi }{2}} \right] = {\textstyle{2 \over 3}}\sin 3\pi\)
:?:
Thank You.
