> Using [[Integral U Substitution]] $\huge \begin{align} \int \tan \pa x \d x &= \int \frac{\sin x}{\cos x} \d x \\ \let u &= \cos x \\ \let \d u &= -\sin x \\ &= \int \frac{\pa{-\frac{\d{u}}{\d x}}}{u}\d x \\ &= -\int u^{-1} \d{u}\\ &= -\ln \left| \cos x \right| + C\\ &= \ln \left| \cos x \right|^{-1} + C\\ &= \ln \left| \frac{1}{\cos x} \right| + C\\ &= \boxed{ \ln \left| \sec x \right| + C} \end{align}$