`int_0^1sec^2 x\ e^(tan\ x)dx`

Let `u=tan\ x`, then `du=sec^2 x\ dx`. So we have:

`int_0^1sec^2x\ e^(tan\ x)dx=[e^(tan x)]_0^1`

`=[e^(tan\ 1)]-[e^(tan\ 0)]`

`=3.747`

Of course, `x` is in radians. These integration techniques don't work in degrees.

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