What is #int 1/(cos(5x-1)) dx#?

1 Answer
Apr 13, 2018

#intdx/cos(5x-1)=1/5ln|sec(5x-1)+tan(5x-1)|+C#

Explanation:

Recalling that #1/cosx=secx,#

#1/cos(5x-1)=sec(5x-1)#

So, we have

#intsec(5x-1)dx#

This is almost a simple integral, just use a quick substitution.

#u=5x-1#

#du=5dx#

#1/5du=dx#

#1/5intsecudu=1/5ln|secu+tanu|+C#

Rewriting in terms of #x# yields

  1. #intdx/cos(5x-1)=1/5ln|sec(5x-1)+tan(5x-1)|+C#