int e^x tan^2(e^x) , dx = | vedclass

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(D) Let $I = \int e^x 	an^2(e^x) \, dx$. Substitute $t = e^x$,then $dt = e^x \, dx$. Substituting these into the integral,we get: $I = \int 	an^2(t)

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$	an(e^x) - e^x + c$

Vedclass · Answer

(D) Let $I = \int e^x 	an^2(e^x) \, dx$. Substitute $t = e^x$,then $dt = e^x \, dx$. Substituting these into the integral,we get: $I = \int 	an^2(t) \, dt$. Using the trigonometric identity $	an^2(t) = \sec^2(t) - 1$,we have: $I = \int (\sec^2(t) - 1) \, dt$. Integrating term by term: $I = 	an(t) - t + c$. Substituting $t = e^x$ back into the expression: $I = 	an(e^x) - e^x + c$.

$\int e^x \tan^2(e^x) \, dx = $

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