Official solution
\[\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} \]
= \[\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}} \]
= \[{1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx\]
\[{1 \over 2}\left[ {{{\tan {x \over 2}} \over {{1 \over 2}}}} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}\]
= \[\left[ {\tan {x \over 2}} \right]_{{\pi \over 4}}^{{3 \over 4}}\]
= tan \[{{3\pi } \over 8}\] - tan\[{\pi \over 8}\]
= \[\left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)\]
= 2