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प्रश्न

\[\int\sec x \cdot \text{log} \left( \sec x + \tan x \right) dx\]
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उत्तर

\[\int\sec x \cdot \log \left( \text{sec x} + \text{tan x} \right) dx\]
\[  \text{Let  log} \left( \sec x + \tan x \right) = t\]
\[ \Rightarrow \frac{\left( \sec x \tan x + \sec^2 x \right)}{\left( \sec x + \tan x \right)} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{\sec x \left( \sec x + \tan x \right)}{\left( \sec x + \tan x \right)} dx = dt\]
\[Now, \int\sec x \cdot \text{log }\left( \sec x + \tan x \right) dx\]
\[ = \ ∫   t  .  dt\]
\[ = \frac{t^2}{2} + C\]
\[ = \frac{\left[ \text{log} \left( \text{sec x} + \tan x \right) \right]^2}{2} + C\]

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पाठ 19: Indefinite Integrals - Exercise 19.09 [पृष्ठ ५८]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.09 | Q 31 | पृष्ठ ५८

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