Question

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

Answer 1

`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))`