Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{1}{{1 - \tan x}}dx = \int {\cfrac{1}{{1 - \cfrac{{\sin x}}{{\cos x}}}}dx} }$

$= \int {\cfrac{{\cos x}}{{\cos x - \sin x}}dx} = \cfrac{1}{2}\int {\cfrac{{2\cos x}}{{{\mathop{\rm cosx}\nolimits} - sinx}}dx}$

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

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

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

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

….(i)
Let ${I_1} = \int {\cfrac{{ - \sin x - \cos x}}{{\cos x - \sin x}}dx}$
Put $\cos x - \sin x = t$ $\Rightarrow$ $\left( { - \sin x - \cos x} \right)dx = dt$

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

….(ii)
From (i) and (ii)

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