Question

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

• 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 $-\frac{1}{8}\cos 4x+C$

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

$=\int \frac{2{\cos }^{2}2x}{{\cos }^{2}x-{\sin }^{2}x}.\sin x\cos xdx$

$2\int \frac{{\cos }^{2}2x.\sin x\cos x}{{\cos }^{2}x}dx$

$\int \cos 2x.\sin 2xdx$

$\frac{1}{2}\int 2\sin 2x\cos 2xdx$

$\frac{1}{2}\int \sin 4xdx$

$-\frac{1}{2}\frac{\cos 4x}{4}+c$

$\frac{-1}{8}\cos 4x+c$