1. We are asked to find the integral of the function $\tan x$. 2. Recall that $\tan x = \frac{\sin x}{\cos x}$. 3. Rewrite the integral as $$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx.$$ 4. Use substitution: let $u = \cos x$, then $du = -\sin x \, dx$ or $-du = \sin x \, dx$. 5. Substitute into the integral: $$\int \frac{\sin x}{\cos x} \, dx = \int \frac{-du}{u} = -\int \frac{1}{u} \, du.$$ 6. Integrate: $$-\int \frac{1}{u} \, du = -\ln |u| + C.$$ 7. Substitute back $u = \cos x$: $$-\ln |\cos x| + C = \ln |\sec x| + C.$$ Final answer: $$\int \tan x \, dx = -\ln |\cos x| + C = \ln |\sec x| + C.$$