$
\eqalign{\int {\frac{{\sin (x)}}
{{\cos (x)}}\,dx = \int {\frac{{ - 1}}
{{\cos (x)}}} } \cdot - \sin (x)\,dx = \int {\frac{{ - 1}}
{u}du = - \ln \left| u \right| + C = - \ln \left| {\cos (x)} \right|} + C}
$
Zie 2. Substitutiemethode
WvR
30-8-2018