Official solution
\[I = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx} \]
\[ = \int\limits_0^{\pi /2} {{{(1 + {{\tan }^2}x/2)dx} \over {3(1 + {{\tan }^2}x/2) + 2(2\tan x/2) + (1 - {{\tan }^2}x/2)}}} \]
Let \[\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dt\]
\[I = \int\limits_0^1 {{{2dt} \over {4 + 2{t^2} + 4t}}} \]
\[ = \int\limits_0^1 {{{dt} \over {{t^2} + 2t + 2}} = \int\limits_0^1 {{{dt} \over {{{(t + 1)}^2} + 1}}} } \]
\[ = \left. {{{\tan }^{ - 1}}(t + 1)} \right|_0^1 = {\tan ^{ - 1}}2 - {\pi \over 4}\]