Answer:
Option A
Explanation:
Let l= $\int \frac{1}{1-\cot x}dx$
= $\int \frac{1}{1-\frac{\cos x}{\sin x}}dx$
= $\int \frac{\sin x}{\sin x-\cos x}dx$
$=\frac{1}{2}\int \frac{2\sin x}{\sin x-\cos x}dx$
$=\frac{1}{2}\int \frac{\sin x+\sin x+\cos x-\cos x}{\sin x-\cos x}dx$
$=\frac{1}{2}\int \frac{(\sin x-\cos x)+(\sin x+\cos x)}{\sin x-\cos x}dx$
$=\frac{1}{2}[\int dx+\int \frac{\sin x+\cos x}{\sin x-\cos x}dx$
Let sin x- cos x=t
$\Rightarrow$ (cos x+ sin x)dx= dt
$\therefore$ $l=\frac{1}{2}\left[\int dx+\int \frac{1}{t}dx\right]$
= $l=\frac{1}{2}\left[x+\log|t|\right]+c$
$=\frac{1}{2}\left[x+\log|\sin x-\cos x|\right]+c$
$\frac{x}{2}+\frac{1}{2}\log|\sin x-\cos x|+c$
Here , A= $\frac{1}{2}$, B= $\frac {1}{2}$
$\therefore$ A+B= $\frac{1}{2}+\frac{1}{2}$=1
