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प्रश्न
\[\int e^x \left( \frac{\sin x \cos x - 1}{\sin^2 x} \right) dx\]
योग
उत्तर
\[\text{ Let I } = \int e^x \left( \frac{\sin x \cos x - 1}{\sin^2 x} \right)dx\]
\[ = \int e^x \left[ \cot x - {cosec}^2 x \right]dx\]
\[\text{ Here}, f(x) = \cot x\]
\[ \Rightarrow f'(x) = - {cosec}^2 x\]
\[\text{ Put e}^x f(x) = t\]
\[ \Rightarrow e^x \cot x = t\]
\[\text{ Diff both sides w . r . t x }\]
\[ e^x \left( \cot x - {cosec}^2 x \right)dx = dt\]
\[ \therefore I = \int dt\]
\[ = t + C\]
\[ = e^x \cot x + C\]
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