Question: How to evaluate the integral:
$$\int \frac{1}{2+\sin x+\cos x}~~dx$$ Let: $$ t＝\tan\left({\frac {x}{2}}\right)\implies\begin{alignedat}{3}\sin(x)＝\frac {2t}{1＋t^2} \\\cos(x)＝\frac {1-t^2}{1＋t^2}\\dx＝\frac {2}{1＋t^2}dt\end{alignedat}\\~\\\int\;\frac {1}{2＋\sin(x)＋\cos(x)}dx＝\int\frac {1}{2＋\frac {2t}{1＋t^2}＋\frac {1-t^2}{1＋t^2}}\frac {1}{1＋t^2}dt\\~\\＝\;\int\;\frac {1}{t^2＋2t＋3}dt$$
How to evaluate : $$\int\;\frac {1}{t^2＋2t＋3}dt$$
$$\frac {1}{t^2＋2t＋3}=\frac{1}{(t+1)^2+2}$$
Let $t+1=\sqrt2 u$
$$\frac {1}{t^2＋2t＋3}=\frac{1}{2(1+u^2)}$$
then
$$\int\;\frac {1}{t^2＋2t＋3}dt=\int\frac{1}{2(1+u^2)}\sqrt2 ~du=\frac{1}{\sqrt2}\arctan u+C=\frac{1}{\sqrt2}\arctan\frac{t+1}{\sqrt2}+C$$