1. Problem: Compute the integral $\int \tan x\, dx$. 2. Formula and rules: Use the identity $\tan x=\frac{\sin x}{\cos x}$ and the substitution rule for integrals. 3. Rewrite the integrand using the identity. $$\int \tan x\, dx = \int \frac{\sin x}{\cos x}\, dx$$ 4. Choose a substitution to simplify the fraction. Let $u=\cos x$ so $du=-\sin x\, dx$. 5. Replace variables in the integral using the substitution. $$\int \frac{\sin x}{\cos x}\, dx = -\int \frac{1}{u}\, du$$ 6. Integrate using the logarithmic rule $\int \frac{1}{u}\, du=\ln|u|+C$. $$-\int \frac{1}{u}\, du = -\ln|u| + C$$ 7. Back-substitute to return to the variable $x$. $-\ln|\cos x| + C$. 8. Final answer: The integral is $$\int \tan x\, dx = -\ln|\cos x| + C = \ln|\sec x| + C$$