Question

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

• A. $\frac{1}{8}\cos 4x+C$
• B. $\frac{1}{4}\cos 4x+C$
• C. $-\frac{1}{4}\cos 4x+C$
• D. $-\frac{1}{8}\cos 4x+C$

Answer 1

The correct option is D $-\frac{1}{8}\cos 4x+C$

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