Question

If $\int \frac{\cos 4x+1}{\cot x-\tan x}dx=A\cos 4x+B$, then

• A. $A=-1/2$
• B. $A=-1/8$
• C. $A=-1/4$
• D. none of these

Answer 1

The correct option is B $A=-1/8$

$\frac{1+\cos 4x}{\cot x-\tan x}=2{\cos }^{2}2x\frac{\tan x}{1-{\tan }^{2}x}={\cos }^{2}2x\tan 2x=\sin 2x\cos 2x=\frac{1}{2}\sin 4x$

$\int \frac{\cos 4x+1}{\cot x-\tan x}dx=\int \frac{1}{2}\sin 4x=-\frac{1}{8}\cos 4x+B$

$⟹A=-\frac{1}{8}$