Question

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

• A. $A=-1/4$ & $B$ may have any value
• B. $A=-1/8$ & $B$ may have any value
• C. $A=-1/2$ & $B=-1/4$
• D. $A=B=1/2$

Answer 1

Answer: (B)

$A=-1/8$ & $B$ may have any value

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

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

$⟹A=\frac{-1}{8}$&$B$ may have any value