Question

$\int \frac{\cos 4x-\cos 2x}{\sin 4x-\sin 2x}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{\cos 4x-\cos 2x}{\sin 4x-\sin 2x}\right)dx \\ =\int \frac{-2\sin \left({\displaystyle \frac{4x+2x}{2}}\right)\sin \left({\displaystyle \frac{4x-2x}{2}}\right)}{2\cos \left({\displaystyle \frac{4x+2x}{2}}\right)\sin \left({\displaystyle \frac{4x-2x}{2}}\right)}dx\left[\because \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\&\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\right] \\ =-\int \frac{\sin 3x}{\cos 3x}dx \\ =-\int \tan 3xdx \\ =\frac{-\ln \left|\sec 3x\right|}{3}+C \\ =\frac{1}{3}\ln {\left(\left|\sec 3x\right|\right)}^{-1}+C \\ =\frac{1}{3}\ln \left|\cos 3x\right|+C \end{gathered}$$