Hi,
u = cos(x) ---> du = -sin(x) dx
$$\int \frac{1}{\sin(x)\cdot u}\cdot\frac{du}{-\sin(x)} = -\int \frac{1}{\sin(x)^2\cdot u} \ du $$
$$-\int\frac{1}{(1-\cos(x)^2)\cdot u}\ du = -\int \frac{1}{u-u^3} \ du = -\ln(u)+\frac12\ln(1-u^2) +c $$
$$ -\ln(\cos(x))+\frac12\ln(1-\cos(x)^2)+c = -\ln(\cos(x))+\ln(\sin(x)) + c$$
Grüße