Evaluate $\int \frac{dx}{1+\sin x+\cos x}$ Evaluate $$\int \frac{1}{1+\sin x+\cos x}\:dx$$
I tried several ways but all of them didn't work
I tried to use Integration-By-Parts method but it's going to give me a more complicated integral
I also tried u-substitution but all of my choices of u didn't work
Any suggestions?

$$\int \frac{1}{1+\sin x+\cos x}\:dx\stackrel{t=\tan(x/2)}=\int\frac{dt}{1+t}=\ln |1+t|+c=\ln|1+\tan(x/2)|+c$$

As $dt=\frac12\sec^2(x/2)dx\implies 2dt=(1+\tan^2(x/2))dx\implies 2dt=(1+t^2)dx$
where:
$$\frac{1}{1+\sin x+\cos x}=\frac{1}{1+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}=\frac{1+t^2}{1+t^2+2t+1-t^2}=\frac{1+t^2}{2t+2}=\frac{\frac{2dt}{dx}}{2(1+t)}$$
