Question

How do you integrate `int e^x sec x dx` using integration by parts?

Answer 1

Let `u = secx` and `dv = e^x dx`

Explanation:

Then

`int e^x secx dx = e^x sec x -int e^x secx tanx dx`

Which WolframAlpha gives as:

`= e^x secx -(1-i)e^((1+i)x) color(white)(l)_2F_1(1/2-i/2,1;3/2-i/2;-e^(2ix))`

Where `color(white)(l)_2F_1(a,b;c;x)` is the hypergeometric function.