Aus Online Mathematik Brückenkurs 2

If we use the definition of \displaystyle \tan x and write the integral as

\displaystyle \int{\tan x\,dx}=\int{\frac{\sin x}{\cos x}\,dx}

we see that the numerator \displaystyle \sin x is the derivative of the denominator (apart from the minus sign). Hence, the substitution \displaystyle u=\cos x will work,

\displaystyle \begin{align} & \int{\frac{\sin x}{\cos x}\,dx}=\left\{ \begin{matrix} u=\cos x \\ du=\left( \cos x \right)^{\prime }\,dx=-\sin x\,dx \\ \end{matrix} \right\} \\ & =-\int{\frac{\,du}{u}}=-\ln \left| u \right|+C=-\ln \left| \cos x \right|+C \\ \end{align}

NOTE: \displaystyle -\ln \left| \cos x \right|+C is only a primitive function in intervals in which \displaystyle \cos x\ne 0.