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