Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 13

$-\frac{\pi }{4}\quad $

\begin{aligned}\int_{\pi / 2}^{0} \frac{1+\cos 2 t}{2} d t&=\frac{1}{2}\int_{\pi / 2}^{0} (1+\cos 2 t)d t\\ &= \frac{1}{2}\left(t+\frac{1}{2}\sin(2t)\right)\bigg|_{\pi / 2}^{0}\\ &= -\frac{1}{2}\frac{\pi}{2}\\ &=-\frac{\pi }{4}\quad \end{aligned}