$1+\cos x = 2\cos^2(x/2)$

$\displaystyle\int_{\pi/4}^{3\pi/4}\dfrac{dx}{2\cos^2(x/2)}=\dfrac{1}{2}\int_{\pi/4}^{3\pi/4}\sec^2\!\left(\dfrac{x}{2}\right)dx=\left[\tan\!\left(\dfrac{x}{2}\right)\right]_{\pi/4}^{3\pi/4}$

$=\tan\!\left(\dfrac{3\pi}{8}\right)-\tan\!\left(\dfrac{\pi}{8}\right)$

Using $\tan(3\pi/8)=\cot(\pi/8)$ and $\cot\theta-\tan\theta=2\cot2\theta$:

$=\cot(\pi/8)-\tan(\pi/8)=2\cot(\pi/4)=2\cdot1=\mathbf{2}$