$\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{-1}\left(\frac{1+\cos 2\theta }{2-\sin 2\theta }\right){\sec }^2\theta d\theta =$ 

$t=\tan \theta \to I=\underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(\frac{1}{1-t+t^2}\right)dt$ 

                          $=\underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(\frac{t+1-t}{1-t(1-t)}\right)dt=\underset{0}{\overset{1}{\int }}\left[{\tan }^{-1}t+{\tan }^{-1}(1-t)\right]dt$

                         $=2\underset{0}{\overset{1}{\int }}{\tan }^{-1}tdt=2\left[{\left[t{\tan }^{-1}t\right]}_0^1-\underset{0}{\overset{1}{\int }}\frac{t}{1+t^2}dt\right]$

                        $=\frac{\pi }{2}{-\left[\ln (1+t^2)\right]}_0^1=\frac{\pi }{2}-\ln 2$