Evaluate definite Integral 1 + sin2x, 0 to pi/2

[val1]

Hi
Can you please check the working out of this solution?

Evaluate the integral and leave pi in your answer.

\(\displaystyle \L\pi \int\limits_0^{{\textstyle{\pi \over 2}}} {(1 + \sin 2x)dx}\)



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

\(\displaystyle \L = \pi \left[ {\frac{\pi }{2} - \frac{1}{2}\cos \pi } \right] - \pi \left[ {0 - \cos 0} \right]\)

\(\displaystyle \L = \pi \left[ {\frac{\pi }{2} + \frac{1}{2}} \right] - \pi \left[ { - 1} \right] = \pi (\frac{\pi }{2} + \frac{3}{2})\)


I think there is a mistake above, but I'm not sure where. Please help.
Thanks

 

[galactus]

Hi
Can you please check the working out of this solution?

Evaluate the integral and leave pi in your answer.

\(\displaystyle \L\pi \int\limits_0^{{\textstyle{\pi \over 2}}} {(1 + \sin 2x)dx}\)



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

\(\displaystyle \L = \pi \left[ {\frac{\pi }{2} - \frac{1}{2}\cos \pi } \right] - \pi \left[ {0 - \cos 0} \right]\)
I believe your problem is in the cos(0). cos(0) is 1. You then have 1/2cos(0)=1/2. See it?.

\(\displaystyle \L = \pi \left[ {\frac{\pi }{2} + \frac{1}{2}} \right] - \pi \left[\underbrace{{ - 1} }\right] = \pi (\frac{\pi }{2} + \frac{3}{2})\)

\(\displaystyle {\pi}\left[\frac{\pi}{2}+\frac{1}{2}\right]-{\pi}[\frac{-1}{2}]=

\frac{{\pi}({\pi}+2)}{2}\)


I think there is a mistake above, but I'm not sure where. Please help.
Thanks