Question

Integrate:

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

• A. $\frac{{\cos }^{2}2x}{4}+\frac{1}{2}\ln \cos 2x+c$
• B. $\frac{{\cos }^{2}2x}{4}+\frac{1}{4}\ln \cos 2x+c$
• C. $\frac{{\cos }^{2}x}{4}-\frac{1}{4}\ln \cos x+c$
• D. $\frac{{\cos }^{2}2x}{4}-\frac{1}{2}\ln \cos 2x+c$

Answer 1

Answer: (D)

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

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

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

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

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

$=\int \frac{\tan 2x(\cos 4x-1)}{2}dx$

$m=2x$ $\Rightarrow dm=2dx$

$\Rightarrow I=\frac{1}{4}\int \frac{\sin m}{\cos m}({2\cos }^{2}m-2)dm$

$\Rightarrow I=\frac{1}{2}\int \cos m\sin mdm-\frac{1}{2}\int \sec m\times \sin mdm$

Let $\cos m=t\Rightarrow \sin mdm=-dt$

$\Rightarrow I=\frac{1}{2}\int tdt-\frac{1}{2}\int \frac{dt}{t}$

$=\frac{t^2}{4}-\frac{1}{2}\ln t+c=\frac{{\cos }^{2}m}{4}-\frac{1}{2}\ln \cos m+c$

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