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∫ E X ( Sin X Cos X − 1 Sin 2 X ) D X - Mathematics

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प्रश्न

\[\int e^x \left( \frac{\sin x \cos x - 1}{\sin^2 x} \right) dx\]
बेरीज
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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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पाठ 19: Indefinite Integrals - Exercise 19.26 [पृष्ठ १४३]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.26 | Q 21 | पृष्ठ १४३

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