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प्रश्न
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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