(C) We know that $\cot x = \frac{\cos x}{\sin x}$.Therefore,$\int \cot x \, dx = \int \frac{\cos x}{\sin x} \, dx$.Let $u = \sin x$,then $du = \cos x

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(C) We know that $\cot x = \frac{\cos x}{\sin x}$.Therefore,$\int \cot x \, dx = \int \frac{\cos x}{\sin x} \, dx$.Let $u = \sin x$,then $du = \cos x \, dx$.Substituting these into the integral,we get $\int \frac{1}{u} \, du = \log |u| + c$.Substituting back $u = \sin x$,we obtain $\log |\sin x| + c$.