Write a value

Let $\log (\sin x)=t$

Differentiating both sides with respect to $x$

$\frac{d t}{d x}=\frac{\cos x}{\sin x} \Rightarrow d t=\frac{\cos x}{\sin x} d x$

$y=\int \frac{1}{t} d t$

Use formula $\int \frac{1}{t} d t=\log t$

$y=\log t+c$

Again, put $t=\log (\sin x)$

$y=\log (\log (\sin x))+c$