int e^x tan x(1+tan x) , dx = . . . . . . | vedclass

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Question

(A) We need to evaluate the integral $I = \int e^x (	an x + 	an^2 x) \, dx$. Recall the identity $1 + 	an^2 x = \sec^2 x$,which implies $	an^2 x =

Vedclass · Accepted Answer

$e^x(	an x - 1)$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int e^x (	an x + 	an^2 x) \, dx$. Recall the identity $1 + 	an^2 x = \sec^2 x$,which implies $	an^2 x = \sec^2 x - 1$. Substituting this into the integral,we get: $I = \int e^x (	an x + \sec^2 x - 1) \, dx$. Rearranging the terms: $I = \int e^x ((	an x - 1) + \sec^2 x) \, dx$. Let $f(x) = 	an x - 1$. Then,$f'(x) = \sec^2 x$. Using the standard integration formula $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$,we have: $I = e^x (	an x - 1) + C$. Therefore,the correct option is $A$.

$\int e^x \tan x(1+\tan x) \, dx = $ . . . . . . $+ C$.

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