Answer
$\frac{\pi}{4}$
Work Step by Step
$\int^{0}_{\frac{\pi}{2}}\frac{1+cos2t}{2}dt$
=$\int^{0}_{\frac{\pi}{2}}(\frac{1}{2}+\frac{1}{2}cos2t)dt$
=$[\frac{1}{2}t+\frac{1}{4}sin2t]^0_{\frac{\pi}{2}}$
=$(\frac{1}{2}0+\frac{1}{4}sin2(0))-(\frac{1}{2}(\frac{\pi}{2})+\frac{1}{4}sin2(\frac{\pi}{2}))$
=$-\frac{\pi}{4}$
