Evaluating a Trigonometric Integral without Substitutions

First express the integrand in the form $$\displaystyle\frac{4\sin 3x \sin 2x \sin x}{4\cos 3x \cos 2x \cos x}.$$ For example, the denominator requires $\cos 2x+\cos 4x\equiv 2\cos 3x \cos x$, $\cos 6x+1\equiv 2\cos^2 3x$, and $\cos 3x+\cos x\equiv 2\cos 2x\cos x$.

Thus the integrand is $\tan 3x\tan 2x\tan x$.

What helps now is the little-known identity $$\tan 3x\tan 2x\tan x\equiv \tan 3x-\tan 2x-\tan x.$$ This starts with $$\sin 3x\sin 2x\sin x \equiv \sin 3x(\cos 2x\cos x-\cos 3x)$$ $$\equiv \sin3x\cos 2x\cos x-\cos3x(\sin 2x\cos x+\cos 2x\sin x),$$ and so on.

Thus the result is $$I=\ln(\cos x)+\frac12\ln(\cos 2x)-\tfrac13\ln(\cos 3x)+c.$$ The identity $\cos x(1-2\cos 2x)\equiv\cos 3x$ implies the equivalence of this answer and the one given above.

The identity $\tan(a+b)\equiv\displaystyle\frac{\tan a+\tan b}{1-\tan a\tan b}$ yields $$\tan a\tan b\tan(a+b)=\tan(a+b)-\tan a-\tan b,$$ so this provides an easier route to the main identity.