$\begin{aligned} & \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1}{2} \sec ^2 \frac{x}{2} d x=\left[\tan \frac{x}{2}\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \\ & =\left(\tan \frac{3 \pi}{8}-\tan \frac{\pi}{8}\right)=\left(\cot \frac{\pi}{8}-\tan \frac{\pi}{8}\right)=\frac{\cos ^2 \frac{\pi}{8}-\sin ^2 \frac{\pi}{8}}{\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}} \\ & =\frac{2 \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}=2\end{aligned}$