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प्रश्न
\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]
बेरीज
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उत्तर
\[\text{Let I }= \int\frac{\ secx \ tanx}{3 \sec x + 5}dx\]
\[\text{Putting }\sec x = t \]
\[ \Rightarrow \frac{dt}{dx} = \sec x \tan x\]
\[ \Rightarrow dt = \text{sec x tan x dx}\]
\[ \therefore I = \int\frac{dt}{3t + 5}\]
\[ = \frac{1}{3} \text{ln }\left| 3t + 5 \right| + C\]
\[ = \frac{1}{3} \text{ln} \left| 3 \sec x + 5 \right| + C \left[ \because t = \sec x \right]\]
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