$$\int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 + \cos \,x}}} $$    is equal to :

Solution :
$$\eqalign{ & I = \int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{1}{2}{{\sec }^2}\frac{x}{2}dx} = \left[ {\tan \frac{x}{2}} \right]_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} \cr & = \tan \frac{{3\pi }}{8} - \tan \frac{\pi }{8} \cr & = \tan \left( {\frac{\pi }{2} - \frac{\pi }{8}} \right) - \tan \frac{\pi }{8} \cr & = \cot \frac{\pi }{8} - \tan \frac{\pi }{8} \cr & = \frac{{{{\cos }^2}\frac{\pi }{8} - {{\sin }^2}\frac{\pi }{8}}}{{\sin \frac{\pi }{8}.\cos \frac{\pi }{8}}} \cr & = \frac{{\cos \frac{\pi }{4}}}{{\frac{1}{2}\sin \frac{\pi }{4}}} \cr & = 2 \cr} $$