Question

If $\int \frac{\cos \left(4x\right)+1}{\cot x-\tan x}dx=f\left(x\right)+C,$ where $C$ is a constant of integration, then $f\left(x\right)$ is

• A. $\frac{1}{8}\cos 4x$
• B. $-\frac{1}{4}\sin 4x$
• C. $\frac{1}{2}\sin 4x$
• D. $-\frac{1}{8}\cos 4x$

Answer 1

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

$I=\int \frac{\cos \left(4x\right)+1}{\cot x-\tan x}dx$
$=\int \frac{2{\cos }^{2}2x}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}dx$
$=\int \frac{2{\cos }^{2}2x}{{\cos }^{2}x-{\sin }^{2}x}\left(\sin x\cos x\right)dx$
$=\int \frac{2{\cos }^{2}2x}{\cos 2x}\left(\sin x\cos x\right)dx$
$=\int \cos 2x\left(2\sin x\cos x\right)dx$
$=\int \frac{1}{2}\sin 4xdx$
$=-\frac{1}{8}\cos 4x+C$