int frac{sec x}{sec x+tan x} dx is equal to | vedclass

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Question

(A) Let $I = \int \frac{\sec x}{\sec x+	an x} dx$. Multiplying the numerator and denominator by $(\sec x - 	an x)$,we get: $I = \int \frac{\sec x(\s

Vedclass · Accepted Answer

$	an x-\sec x+C$

Vedclass · Answer

(A) Let $I = \int \frac{\sec x}{\sec x+	an x} dx$. Multiplying the numerator and denominator by $(\sec x - 	an x)$,we get: $I = \int \frac{\sec x(\sec x - 	an x)}{\sec^2 x - 	an^2 x} dx$. Since $\sec^2 x - 	an^2 x = 1$,the integral simplifies to: $I = \int (\sec^2 x - \sec x 	an x) dx$. Integrating term by term,we obtain: $I = 	an x - \sec x + C$.

$\int \frac{\sec x}{\sec x+\tan x} dx$ is equal to

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