Let $I=\int \cot x \cdot \log [\log (\sin x)] d x$
Put $\begin{aligned} \log (\sin x) &=t \Rightarrow \frac{\cos x}{\sin x} d x=d t \Rightarrow \cot x d x=d t \\ \therefore I &=\int \log t d t=\int \log t \cdot 1 d t \\ &=\log t \int 1 d t-\int\left[\frac{d}{d t}(\log t) \cdot \int 1 d t\right] d t \\ &=(\log t) \cdot t-\int \frac{1}{t} \cdot t d t=t \cdot \log t-\int 1 d t \\ &=t \cdot \log t-t+c=t(\log t-1)+c \\ &=\log (\sin x)[\log (\log (\sin x))-1]+c \end{aligned}$