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