Question

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

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

Answer 1

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

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

$=\int \frac{1+2{\cos }^2\left(2x\right)-1}{\frac{\cos x}{\sin x}-\frac{sinx}{\cos x}}dx$

$=\int \frac{2{\cos }^2\left(2x\right)}{\frac{{\cos }^{2}x-{\sin }^{2}x}{\sin x.\cos x}}dx$

$=\int \frac{2{\cos }^2\left(2x\right)}{\frac{\cos 2x}{\sin x.\cos x}}dx$

$=\int 2.\cos 2x.(\sin x.\cos x)dx$

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

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

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