Let $I=\int\frac{1}{1-\tan x}dx$ $=\int\frac{1}{1-\frac{\sin x}{\cos x}}dx$ $=\int\frac{\cos x}{\cos x-\sin x}dx$ $=\frac{1}{2}\int\frac{2\cos x}{\cos x-\sin x}dx$ $=\frac{1}{2}\int1dx-\frac{1}{2}\int\frac{-\sin x-\cos x}{\cos x-\sin x}dx$ $=\frac{x}{2}-\frac{1}{2}\int\frac{-\sin x-\cos x}{\cos x-\sin x}dx+C_1$ $I=\frac{x}{2}-\frac{1}{2}I_1+C_1$ ... (i) where, $I_1=\int\frac{-\sin x-\cos x}{\cos x-\sin x}dx$Put $\cos x - \sin x = t$ $\Rightarrow (-\sin x - \cos x) dx = dt$$I_1=\int\frac{dt}{t}=\log|t|+C_2$ $=\log|\cos x-\sin x|+C_2$ ... (ii)From (i) and (ii), we get $I=\frac{x}{2}-\frac{1}{2}\log|\cos x-\sin x|+C$