Integrate the function $\cot x \log \sin x$.
(N/A) Let $I = \int \cot x \log \sin x \, dx$.
Substitute $t = \log \sin x$.
Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Therefore,$dt = \cot x \, dx$.
Substituting these into the integral,we get $I = \int t \, dt$.
Integrating with respect to $t$,we get $I = \frac{t^2}{2} + C$.
Substituting back $t = \log \sin x$,we get $I = \frac{1}{2}(\log \sin x)^2 + C$,where $C$ is an arbitrary constant.
Substitute $t = \log \sin x$.
Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Therefore,$dt = \cot x \, dx$.
Substituting these into the integral,we get $I = \int t \, dt$.
Integrating with respect to $t$,we get $I = \frac{t^2}{2} + C$.
Substituting back $t = \log \sin x$,we get $I = \frac{1}{2}(\log \sin x)^2 + C$,where $C$ is an arbitrary constant.
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