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?

If you want to avoid the Weierstrass substitution ($t = \tan(\frac{x}{2})$), you can multiply the top and bottom by
$1+ \sin(x) - \cos(x)$.
$$\int \frac{dx}{1+\sin(x) + \cos(x)} = \int \frac{(1 + \sin(x) - \cos(x))dx}{1 + 2\sin(x) + \sin^2(x) - \cos^2(x)} $$
$$= \int \frac{1 + \sin(x)dx}{2\sin(x) + 2\sin^2(x)} - \int \frac{\cos(x)dx}{2\sin(x) + 2\sin^2(x)} $$
$$= \int {\frac{\csc(x)dx}{2}}- \frac{1}{2}\int \Big(\frac{\cos(x)}{\sin(x)} - \frac{\cos(x)}{(1 + \sin(x))}\Big)dx = \cdots$$
The last two integrals can be solved using the methods you already know.
