Question

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

Answer 1

\[\int\left( \frac{\cos4x - \cos2x}{\sin4x - \sin2x} \right)dx\]
\[ = \int\frac{- 2\sin\left( \frac{4x + 2x}{2} \right)\sin\left( \frac{4x - 2x}{2} \right)}{2\cos\left( \frac{4x + 2x}{2} \right)\sin\left( \frac{4x - 2x}{2} \right)}dx \left[ \because \cos A - \cos B = - 2\sin \left( \frac{A + B}{2} \right)\sin \left( \frac{A - B}{2} \right) \text{and} \sin A - \sin B = 2\cos \left( \frac{A + B}{2} \right)\sin \left( \frac{A - B}{2} \right) \right]\]
\[ = - \int\frac{\sin 3x}{\cos 3x}dx\]
\[ = - \int\tan 3x dx\]
\[ = \frac{- \text{ln }\left| \sec 3x \right|}{3} + C\]
\[ = \frac{1}{3} \text{ln} \left( \left| \text{sec 3x} \right| \right)^{- 1} + C\]
\[ = \frac{1}{3} \text{ln }\left| \cos 3x \right| + C\]