Question

${\int }_{}^{}\frac{\cos 4x-1}{\cot x-\tan x}dx$ is equal to:

• A. $-\frac{1}{2}\cos 4x+c$
• B. $-\frac{1}{4}\cos 4x+c$
• C. $-\frac{1}{2}\sin 2x+c$
• D. none of these

Answer 1

Answer: (D)

Let $I=\int \frac{\cos 4x-1}{\cot x-\tan x}dx$
$=-\int \frac{2{\sin }^{2}2x}{\frac{{\cos }^{2}x-{\sin }^{2}x}{\cos x\sin x}}dx$
$=-\int \frac{\sin 2x\left(1-{\cos }^{2}2x\right)}{\cos 2x}dx$ Put $\cos 2x=t$
$\Rightarrow -2\sin 2xdx=dt$
$\therefore I=\frac{1}{2}\int \frac{\left(1-t^2\right)}{t}dt$
$=\frac{1}{2}\left[\int \left(\frac{1}{2}-t\right)dt\right]$
$=\frac{1}{2}\left[\log t-\frac{t^2}{2}\right]+c$
$=\frac{1}{2}\log \cos 2x-\frac{{\cos }^{2}2x}{4}+c$