If $\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x=A \sqrt{\cot x}+B$, then A is equal to

1 2 -1 -2

-2

$\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x=\int \frac{\sqrt{\cot x}}{\cot x} . ~cosec^2 x~dx=\int \frac{cosec^2 x}{\sqrt{\cot x}} d x=-2 \sqrt{\cot x}+B$ Hence A = –2. Hence (4) is the correct answer.