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Mathematics 8 Online
OpenStudy (anonymous):

can somone help me evaluate this integral

OpenStudy (anonymous):

OpenStudy (anonymous):

You have: \[\int\limits_{\frac{\pi}{2}}^{0} \frac{1+\cos(2t)}{2}dt=\frac{1}{2} \int\limits_{\frac{\pi}{2}}^{0}1+\cos(2t)dt\] For the second part use u=2t du=2dt (1/2)du=dt Replacing this you have: \[\frac{1}{2}\int\limits_{\frac{\pi}{2}}^{0}1dt+\frac{1}{2}(\frac{1}{2})\int\limits \cos(u)du=\frac{1}{2}t|_{\frac{\pi}{2}}^{0}+\frac{1}{4}\sin(u)\] Replugging in your u you have: \[[\frac{t}{2}+\frac{1}{4}\sin(2t)]_{\frac{\pi}{2}}^{0}=(0+0)-(\frac{\pi}{4}+\frac{1}{4}(0))=-\frac{\pi}{4}\]

OpenStudy (anonymous):

thanks

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