Getting to Grips with Intergrating Trig Functions 3

[Guest]

Last one guys...

\(\displaystyle \L
\int_0^\pi {2\sin (2x + \frac{\pi }{2}} )dx\)

\(\displaystyle \L

= - {\textstyle{1 \over 2}}\left[ {\cos (2x + \frac{\pi }{2})} \right]_0^\pi = - {\textstyle{1 \over 2}}\left[ {\cos (2\pi + \frac{\pi }{2}) - \cos (2(0) + \frac{\pi }{2})} \right]\)

\(\displaystyle \L

= - {\textstyle{1 \over 2}}\left[ {\cos \frac{{5\pi }}{2} - \cos \frac{\pi }{2}} \right] = - {\textstyle{1 \over 2}}\cos 2\pi\)

:?: Thanks a lot.

 

[skeeter]

same mistake as your other post ... sin(a) - sin(b) does not equal sin(a-b).

 

[tkhunny]

\(\displaystyle cos(\frac{\pi}{2})\,=\,cos(\frac{5\pi}{2})\,=\,0\)

This may not be the intent of the problem. It is always a good idea to have some idea of that with which you are dealing. Things that cross the x-axis can be a little difficult to understand.

 

[Guest]

Thanks for your help.

I've got another one I'm trying.
Let's see if I can get that one right...