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

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Answer 1 · 2016-12-14T18:41:55

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

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.

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