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∫ E X ( Tan X − Log Cos X ) D X - Mathematics

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प्रश्न

\[\int e^x \left( \tan x - \log \cos x \right) dx\]
योग

उत्तर

\[\text{ Let I } = \int e^x \left( \tan x - \log \cos x \right)dx\]

\[\text{ here }f(x) = - \text{ log 
}\text{ cos x Put e} ^x f(x) = t\]

\[ \Rightarrow f'(x) = \tan x\]

\[\text{let - e}^x \text{ log }\text{ cos x } = t\]

\[\text{ Diff both  sides  w . r . t x }\]

\[ - \left[ e^x \text{ log }\left( \text{ cos x } \right) + e^x \frac{1}{\cos x} \times \left( - \sin x \right) \right] = \frac{dt}{dx}\]

\[ \Rightarrow \left[ - e^x \text{ log }\left( \text{ cos x } \right) + e^x \tan x \right]dx = dt\]

\[ \therefore \int e^x \left( \tan x - \text{ log }\cos x \right)dx = \int dt\]

\[ = t + C\]

\[ = - e^x \text{ log }\left( \text{ cos x }\right) + C\]

\[ = e^x \text{ log }\left( \sec x \right) + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.26 [पृष्ठ १४३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.26 | Q 7 | पृष्ठ १४३

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