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प्रश्न
\[\int\frac{sec x}{\log \left( \text{sec x }+ \text{tan x} \right)} dx\]
बेरीज
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उत्तर
` Note: "Here, we are considering "log x as log_e x`
\[\text{Let I} = \int\frac{\sec x}{\log \left( \sec x + \tan x \right)}dx\]
\[\text{Putting} \log \left( \sec x + \tan x \right) = t\]
\[ \Rightarrow \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \frac{dt}{dx}\]
\[ \Rightarrow \sec x\frac{\left( \sec x + \tan x \right)}{\sec x + \tan x} = \frac{dt}{dx}\]
\[ \Rightarrow \text{sec x dx} = dt\]
\[ \therefore I = \int\frac{dt}{t}\]
\[ = \text{log}\left| t \right| + C\]
\[ = \text{log}\left| \text{log }\left( \sec x + \tan x \right) \right| + C\]
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