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प्रश्न
\[\int e^x \left( \cot x + \log \sin x \right) dx\]
बेरीज
उत्तर
\[\text{ Let I }= \int e^x \left( \cot x + \log \sin x \right)dx\]
\[\text{ Here,} f(x) = \log \sin x Put e^x f(x) = t\]
\[ \Rightarrow f'(x) = \cot x\]
\[\text{ let e}^x \log \sin x = t\]
\[\text{ Diff both sides w . r . t x}\]
\[ e^x \text{ log } \left( \sin x \right) + e^x \times \frac{1}{\sin x} \times \cos x = \frac{dt}{dx}\]
\[ \Rightarrow \left[ e^x \text{ log}\left( \sin x \right) + e^x \cot x \right]dx = dt\]
\[ \Rightarrow e^x \left( \cot x + \text{ log }\sin x \right)dx = dt\]
\[ \therefore \int e^x \left( \cot x + \log \sin x \right)dx = \int dt\]
\[ = t + C\]
\[ = e^x \log \sin x + C\]
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