Question

Integrate the following functions.
$\int \frac{1}{(1-tanx)}dx.$

Answer 1

Let $I=\int \frac{1}{(1-tanx)}dx=\int \frac{1}{1-\frac{sinx}{cosx}}dx=\int \frac{1}{\frac{cosx-sinx}{cosx}}dx$
$=\frac{1}{2}\int \frac{2\left(cosx\right)dx}{(cosx-sinx)}=\frac{1}{2}\int \frac{cosx+cosx+sinx-sinx}{(cosx-sinx)}dx$
[add and subtract cos x in numerator]
$=\frac{1}{2}[\int \frac{(cosx-sinx)}{(cosx-sinx)}dx+\int \frac{(cosx+sinx)}{(cosx-sinx)}dx]=\frac{1}{2}[\int 1dx+\int \frac{(cosx+sinx)}{(cosx-sinx)}dx]$

Let cos x -sin x=t
$\Rightarrow -sinx-cosx=\frac{dt}{dx}\Rightarrow -[sinx+cosx]=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{-[sinx+cosx]}\therefore I=\frac{1}{2}[\int 1dx+\int \frac{cosx+sinx}{t}\frac{dt}{-[sinx+cosx]}]=\frac{1}{2}[\int 1dx-\int \frac{1}{t}dt]=\frac{1}{2}[x-log|t|]+C=\frac{1}{2}[x-log|cosx-sinx|]+C$