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