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प्रश्न
उत्तर
\[\text{Note: Here we are considering} \log\ x\ as \log_e x\]
\[\text{Let I} = \int\frac{\cot x}{\log \sin x}dx\]
\[\text{Putting} \log \sin x = t\]
\[ \Rightarrow \cot x = \frac{dt}{dx}\]
\[ \Rightarrow \text{cot x dx }= dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{log }\left| t \right| + C \]
\[ = \text{log }\left| \log \sin x \right| + C \left[ \because t = \log \sin x \right]\]
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