$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\sin x+\cos x}{3+\sin 2x}dx=\_\_\_$

The integral can be written as 

$-\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\sin x+\cos x}{{\left(\sin x-\cos x\right)}^2-4}dx$

Put $t=\sin x-\cos x$ then 

$dt=\left(\cos x+\sin x\right)dx$

$-\underset{-1}{\overset{0}{\int }}\frac{dt}{t^2-4}=$$-\underset{\theta -1}{\overset{\theta }{\int }}\frac{dt}{t^2-2^2}$$=-\frac{1}{4}{\left[\log \left|\frac{t-2}{t+1}\right|\right]}_{-1}^0$

$=-\frac{1}{4}\left(\log 1-\log 3\right)=\frac{1}{4}\log 3$