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