<s>[INST] Integrate $\int \frac{1}{1+ \tan x}dx$ Does this integral have a closed form?
$$\int \frac{1}{1+ \tan x}\,dx$$
My attempt:
$$\int \frac{1}{1+ \tan x}\,dx=\ln (\sin x + \cos x) +\int \frac{\tan x}{1+ \tan x}\,dx$$
What is next?
 [/INST] $$
I=\int  \frac 1 {1+\tan  x} \, dx=\int \frac {\cos x}{\cos x +\sin x} \, dx\\
J=\int \frac {\sin x}{\cos x +\sin x} \, dx\\
I+J=\int \frac {\cos x+\sin x}{\cos x +\sin x} \, dx = x+C_1\\ \\
I-J=\int \frac {\cos x-\sin x}{\cos x +\sin x} \,dx=\int \frac { d\left( \sin x +\cos x\right)  }{ \cos x +\sin x} =\ln \left| \sin x +\cos x \right| + C_2 \\
2I = x+\ln \left| \sin x +\cos x\right| + C_3 \\
I=\frac 1 2 \left[ x+\ln \left| \sin x +\cos x \right| + C_3 \right]
$$
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