How do you integrate #e^(tanx) * sec^2(x) dx#?

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Answer 1 · 2016-07-12T04:33:25

#int e^(tan x) sec^(2)(x) dx = e^(tan x) + C#

We can use a simple u-substitution to evaluate this integral.

Given #int e^(tan x) sec^(2)(x) dx#

Note that #sec^2(x)# is the derivative of #tan x#, thus we can use

#u = tan x -> du = sec^(2)(x) dx#

Since

#int e^(u) du = e^(u) + C#

We only have to substitute #u# back into our result to get

#int e^(tan x) sec^(2)(x) dx = e^(tan x) + C#