Evaluate the following integrals: Evaluate: . Hence, evaluate: .

$\int \tan x dx = \int \frac{\sin x}{\cos x} dx = -\int \frac{-\sin x}{\cos x} dx = -\ln|\cos x| + C = \ln|\sec x| + C$.

We rewrite the integrand: $I = \int \frac{dx}{\cot\frac{x}{2} \cot\frac{x}{3} \cot\frac{x}{6}} = \int \tan\frac{x}{2} \tan\frac{x}{3} \tan\frac{x}{6} dx$. Notice that $\frac{x}{2} = \frac{x}{3} + \frac{x}{6}$. Taking tan on both sides: $\tan\frac{x}{2} = \tan\left(\frac{x}{3} + \frac{x}{6}\right) = \frac{\tan\frac{x}{3} + \tan\frac{x}{6}}{1 - \tan\frac{x}{3}\tan\frac{x}{6}}$. Cross-multiplying: $\tan\frac{x}{2}\left(1 - \tan\frac{x}{3}\tan\frac{x}{6}\right) = \tan\frac{x}{3} + \tan\frac{x}{6}$ $\tan\frac{x}{2} - \tan\frac{x}{2}\tan\frac{x}{3}\tan\frac{x}{6} = \tan\frac{x}{3} + \tan\frac{x}{6}$ $\tan\frac{x}{2}\tan\frac{x}{3}\tan\frac{x}{6} = \tan\frac{x}{2} - \tan\frac{x}{3} - \tan\frac{x}{6}$. Integrating each term using $\int \tan(kx) dx = -\frac{1}{k}\ln|\cos(kx)|$: $I = \int \tan\frac{x}{2} dx - \int \tan\frac{x}{3} dx - \int \tan\frac{x}{6} dx$ $= -2\ln\left|\cos\frac{x}{2}\right| - (-3)\ln\left|\cos\frac{x}{3}\right| - (-6)\ln\left|\cos\frac{x}{6}\right| + C$ $= -2\ln\left|\cos\frac{x}{2}\right| + 3\ln\left|\cos\frac{x}{3}\right| + 6\ln\left|\cos\frac{x}{6}\right| + C$.