$\cfrac{1}{{1 + \cot x}}$

: Let$I = \int {\cfrac{1}{{1 + \cot x}}dx} = \int {\cfrac{1}{{1 + \cfrac{{\cos x}}{{\sin x}}}}dx} = \int {\cfrac{{\sin x}}{{\sin x + \cos x}}dx}$

$= \cfrac{1}{2}\int {\cfrac{{2\sin x}}{{\sin x + \cos x}}} dx$

$= \cfrac{1}{2}\int {\cfrac{{\left( {\sin x + \cos x} \right) - \left( {\cos x - \sin x} \right)}}{{\left( {\sin x + \cos x} \right)}}dx}$

$= \cfrac{1}{2}\int {\left( 1 \right)dx} - \cfrac{1}{2}\int {\cfrac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx$

$= \cfrac{1}{2}x - \cfrac{1}{2}\int {\cfrac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx + {C_1}$

$\therefore$ $I = \cfrac{x}{2} - \cfrac{1}{2}{I_1} + {C_1}$

…..(i)
Let ${I_1} = \int {\cfrac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx$

Put $\sin x + \cos x = t$ $\Rightarrow$ $\left( {\cos x - \sin x} \right)dx = dt$

$\Rightarrow$ ${I_1} = \int {\cfrac{{dt}}{t}} = \log \left| t \right| + {C_2} = \log \left| {\cos x + \sin x} \right| + {C_2}$

…..(ii)
From (i) and (ii),

we get
$\Rightarrow$ $I = \cfrac{1}{2}x - \cfrac{1}{2}\log \left| {\cos x + \sin x} \right| + C$
Where, $C = {C_1} + {C_2}$