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

$$\begin{gathered} \underset{0}{\overset{\mathrm{\pi }/4}{\int }}\frac{\mathrm{sinx}+\mathrm{cosx}}{3+\sin 2\mathrm{x}}\mathrm{dx}= \\ \mathrm{Put} \mathrm{Sin} \mathrm{x}-\cos \mathrm{x}=\mathrm{t} \\ (\cos \mathrm{x}+\sin \mathrm{x})\mathrm{dx}=\mathrm{dt} \\ \begin{array}{c}{\int }_1^0 \frac{\mathrm{dt}}{4-{\mathrm{t}}^2}\\ \frac{1}{2\left(2\right)}\log {\left|\frac{2+\mathrm{t}}{2-\mathrm{t}}\right|}_{-1}^0\end{array} \overline{) \begin{array}{c}{\mathrm{t}}^2=Si{n}^2 \mathrm{x}+{\cos }^2\mathrm{x}-2Sin \mathrm{x} \cos \mathrm{x}\\ =1-\mathrm{Sin} 2\mathrm{x}\\ \mathrm{Sin} 2\mathrm{x}=1-{\mathrm{t}}^2\end{array}} \\ =\frac{1}{4}\left[\log 1-\log \left(\frac{1}{3}\right)\right] \\ =\frac{1}{4}[\log 3] \end{gathered}$$