## University Calculus: Early Transcendentals (3rd Edition)

$\ln |\sec a+\tan a|+c=\ln |\sec e^t +\tan e^t|+c$
Consider the integral $\int e^t \sqrt {\tan^2 e^t+1}dt$ Plug in $a=e^t \implies e^t dt =da$ The given integral can be re-written as: $\int \sqrt {\tan^2 a+1}dt=\int \sqrt {sec^2 a} da$ This implies that $\sec (a) da=\ln |\sec a+\tan a|+c=\ln |\sec e^t +\tan e^t|+c$