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

1 Answer
Dec 14, 2016

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.