If $\int \frac{1}{1+\cot x}dx=Ax+B\log |\cos x+\sin x|+c$ then (A, B) =

$I=\int \frac{1}{1+\cot x}dx$

$=\int \frac{\sin x}{\sin x+\cos x}dx$

$=\int \frac{\cos x+\sin x}{\cos x+\sin x}dx$

Formula

$=\int \frac{a\cos x+b\sin x}{c\cos x+d\sin x}dx=ax+b\log |c\cos x+d\sin x|+c$

$A=\frac{ac+bd}{c^2+d^2}, B=\frac{ad-bc}{c^2+d^2}$

$=\frac{0\left(1\right)+1\left(1\right)}{1^2+1^2}, \frac{0\left(1\right)-1\left(1\right)}{1^2+1^2}$

$=\frac{1}{2}, =\frac{-1}{2}$

$I=\frac{1}{2}x-\frac{1}{2}\log |\cos x+\sin x|+c$

$(A,B)=\left(\frac{{\displaystyle 1}}{{\displaystyle 2}},\frac{{\displaystyle -1}}{{\displaystyle 2}}\right)$