How do you integrate #int e^(sec2x)sec2xtan2xdx# from #[pi/3,pi/2]#?

1 Answer
Apr 21, 2017

Answer:

#1/2 * (e^(-1) - e^(-2))#

Explanation:

#d/dx(sec 2x) = d/dx (1/(cos 2x)) = - 1/(cos 2x)^2 * (-sin 2x) * 2#
#= sec 2x * tan 2x * 2#

Hence

#d/dx e^(sec 2x) = e^(sec 2x) * d/dx(sec 2x) = e^(sec 2x) * sec 2x * tan 2x * 2#

So

#int_(pi/3)^(pi/2) e^(sec2x)sec2xtan2xdx = 1/2 e^(sec 2x)|_(pi/3)^(pi/2) = 1/2 * (e^(-1) - e^(-2))#