int frac{tan x}{sec x + tan x} , dx = | vedclass

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Question

(B) We have the integral $I = \int \frac{	an x}{\sec x + 	an x} \, dx$.Multiplying the numerator and denominator by $(\sec x - 	an x)$,we get:$I =

Vedclass · Accepted Answer

$\sec x - 	an x + x + c$

Vedclass · Answer

(B) We have the integral $I = \int \frac{	an x}{\sec x + 	an x} \, dx$.Multiplying the numerator and denominator by $(\sec x - 	an x)$,we get:$I = \int \frac{	an x(\sec x - 	an x)}{(\sec x + 	an x)(\sec x - 	an x)} \, dx$Since $(\sec^2 x - 	an^2 x) = 1$,the expression simplifies to:$I = \int (\sec x 	an x - 	an^2 x) \, dx$Using the identity $	an^2 x = \sec^2 x - 1$,we have:$I = \int (\sec x 	an x - (\sec^2 x - 1)) \, dx$$I = \int \sec x 	an x \, dx - \int \sec^2 x \, dx + \int 1 \, dx$Integrating each term,we get:$I = \sec x - 	an x + x + c$.

$\int \frac{\tan x}{\sec x + \tan x} \, dx = $

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