\(\int \frac{d x}{\sin x+\cos x}=\)

\(\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{\mathrm{d} x}{\sin x+\cos x} \\ & \text { Put } \tan \frac{x}{2}=\mathrm{t} \\ & \Rightarrow x=2 \tan ^{-1} \mathrm{t} \\ & \Rightarrow \mathrm{d} x=\frac{2}{1+\mathrm{t}^2} \mathrm{dt} \\ & \text { and } \cos x=\frac{1-t^2}{1+t^2}, \sin x=\frac{2 t}{1+t^2} \\ & \therefore \quad I=\int \frac{\frac{2}{1+t^2}}{\frac{2 t}{1+t^2}+\frac{1-t^2}{1+t^2}} d t \\ & =\int \frac{2}{2 \mathrm{t}+1-\mathrm{t}^2} d \mathrm{t} \\ & =-2 \int \frac{1}{\mathrm{t}^2-2 \mathrm{t}-1} d \mathrm{t} \\ & =-2 \int \frac{1}{t^2-2 t+1-1-1} \\ & =-2 \int \frac{1}{(t-1)^2-(\sqrt{2})^2} \\ & \end{aligned}\)
\(\begin{aligned} & =-2 \times \frac{1}{2 \sqrt{2}} \log \left|\frac{\mathrm{t}-1-\sqrt{2}}{\mathrm{t}-1+\sqrt{2}}\right|+\mathrm{c} \\ & =\frac{-1}{\sqrt{2}} \log \left|\frac{\tan \frac{x}{2}-1-\sqrt{2}}{\tan \frac{x}{2}-1+\sqrt{2}}\right|+\mathrm{c} \\ & =\frac{-1}{\sqrt{2}} \log \left|\frac{\tan \frac{x}{2}-(\sqrt{2}+1)}{\tan \frac{x}{2}+\sqrt{2}-1}\right|+\mathrm{c}\end{aligned}\)