Question

Prove that $\int \frac{\cot x-\tan x}{\cos 4x+1}dx$ is equal to $\frac{1}{2}\ln |\tan 2x|+C$, where $C$ is an integration constant.

Answer 1

$f\left(x\right)=\int \frac{\cot x-\tan x}{\cos 4x+1}dx=\int \frac{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}{2{\cos }^{2}2x-1+1}dx(\because 2{\cos }^{2}2x-1=\cos 4x)$

$=\int \frac{\frac{{\cos }^{2}x-{\sin }^{2}x}{\sin x.\cos x}}{2{\cos }^{2}2x}dx(\because \cos 2x={\cos }^{2}x-{\sin }^{2}x)$
$=\int \frac{\cos 2x}{2\sin x.\cos x.{\cos }^{2}2x}dx$
$=\int \frac{1}{\sin 2x.\cos 2x}dx$

On multiplying and dividing the above expression by $2$, we have
$f\left(x\right)=\int \frac{2}{\sin 4x}dx$ $(\because 2\sin 2x\cos 2x=\sin 4x)$
$=2\int \text{cosec}4xdx$
$=\frac{2}{4}\ln |\text{cosec}4x-\cot 4x|+C$
$=\frac{1}{2}\ln |\frac{1}{\sin 4x}-\frac{\cos 4x}{\sin 4x}|+C$
$=\frac{1}{2}\ln \left|\frac{1-\cos 4x}{\sin 4x}\right|+C$
$=\frac{1}{2}\ln \left|\frac{2{\sin }^{2}2x}{2\sin 2x\cdot \cos 2x}\right|+C$
$=\frac{1}{2}\ln |\tan 2x|+C$ (proved)