How do you find the antiderivative of $e^{x} (1 - (e^{- x}) {sec}^{2} x) d x$ $e^x(1-(e^-x)sec^2x) dx$ ?

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Answer 1 · 2017-11-15T22:38:15

$\int e^{x} (1 - e^{- x} {sec}^{2} x) d x = e^{x} - tan x + C$ $int e^x(1-e^(-x)sec^2x) dx = e^x-tanx+C$

Simplify the expression:

$e^{x} (1 - e^{- x} {sec}^{2} x) = e^{x} - {sec}^{2} x$ $e^x(1-e^(-x)sec^2x) = e^x -sec^2x$

then using the linearity of the integral:

$\int e^{x} (1 - e^{- x} {sec}^{2} x) d x = \int e^{x} d x - \int {sec}^{2} x d x$ $int e^x(1-e^(-x)sec^2x) dx = int e^xdx -int sec^2xdx$

Both terms are standard integrals:

$\int e^{x} d x = e^{x} + c_{1}$ $int e^xdx = e^x+c_1$

$\int {sec}^{2} x d x = tan x + c_{2}$ $int sec^2xdx = tanx +c_2$

and then:

$\int e^{x} (1 - e^{- x} {sec}^{2} x) d x = e^{x} - tan x + C$ $int e^x(1-e^(-x)sec^2x) dx = e^x-tanx+C$

How do you find the antiderivative of ex(1−(e−x)sec2x)dxe^x(1-(e^-x)sec^2x) dx?