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
Answer & explanation
Correct answer: option 2
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 $B \in R$