Evaluate $\int \cos 2x \; \cos 4 x \; \cos 6x \; dx$
$\begin{aligned} \text{Let } I & = \int \cos 2x \; \cos 4x \; \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int 2 \cos 2x \; \cos 4x \; \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int \left\{\cos \left(4x + 2x\right) + \cos \left(4x - 2x\right) \right\} \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int \left(\cos 6x + \cos 2x\right) \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int \cos^2 6x \; dx + \dfrac{1}{2} \int \cos 2x \; \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int \dfrac{1 + \cos 12x}{2} \; dx + \dfrac{1}{4} \int 2 \; \cos 2x \; \cos 6x \; dx \\\\ & = \dfrac{1}{4} \int dx + \dfrac{1}{4} \int \cos 12x \; dx + \dfrac{1}{4} \int \cos \left(6x+2x\right) + \cos \left(6x - 2x\right) \; dx \\\\ & = \dfrac{1}{4} \left\{\int dx + \int \cos 12x \; dx + \int \cos 8x \; dx + \int \cos 4x \; dx \right\} \\\\ & = \dfrac{1}{4} \left\{x + \dfrac{\sin 12x}{12} + \dfrac{\sin 8x}{8} + \dfrac{\sin 4x}{4} \right\} + c \\\\ & = \dfrac{x}{4} + \dfrac{\sin 12x}{48} + \dfrac{\sin 8x}{32} + \dfrac{\sin 4x}{16} + c \\\\ & \left[\begin{aligned} \text{Note: } & 2 \cos A \; \cos B = \cos \left(A + B\right) + \cos \left(A - B\right) \\\\ & \cos^2 \theta = \dfrac{1 + \cos 2 \theta}{2} \end{aligned}\right] \end{aligned}$