If $\int \frac{1+\cos 4 x}{\cot x-\tan x} d x=A \cos 4 x+B$, then the values of A and B are

$A=\frac{1}{8}, B \in R$ $A=-\frac{1}{8}, B \in R$ $A=\frac{1}{4}, B \in R$ none of these

$A=-\frac{1}{8}, B \in R$

We have, $I=\int \frac{1+\cos 4 x}{\cot x-\tan x} d x$ $\Rightarrow I =\int \frac{2 \cos ^2 2 x \sin x \cos x}{\cos ^2 x-\sin ^2 x}=\int \sin 2 x \cos 2 x d x$ $\Rightarrow I =\frac{1}{2} \int \sin 4 x d x=-\frac{1}{8} \cos 4 x+B$ Hence, $A=-\frac{1}{8}$ and $B \in R$