Expert: Paul Klarreich - 7/30/2009

Question
I have a question about the Integration of Sec(x)dx which is: ln|sec(x)+tan(x)|
sec(x)=sec(-x) for -π/2<x<π/2, but ln|Sec(x)+Tan(x)| != ln|Sec(-x)+Tan(-x)|
so integrating for example sec(x) from 0 to π/4 (= ln|Sec(π/4)+Tan(π/4)|-ln|Sec(0)+Tan(0)| = 0.693-0 = 0.693) gives a different result from integrating sec(x) from -π/4 to 0 (= ln|Sec(0)+Tan(0)-ln|Sec(-π/4)+Tan(-π/4)| = 0-( -36.737)= 36.737).
Why is this, is it because this result is only viable for 0<x<π/2? If so, can you explain it?

Answer
Questioner: jan
Country: Netherlands
Category: Calculus
Private: No
Subject: Integration of Sec(x)dx
Question: I have a question about the Integration of Sec(x)dx which is:

ln|sec(x)+tan(x)|

sec(x)=sec(-x) for -π/2<x<π/2,
but ln|Sec(x)+Tan(x)| != ln|Sec(-x)+Tan(-x)|

so integrating for example sec(x) from 0 to π/4  
= ln|Sec(π/4)+Tan(π/4)| - ln|Sec(0)+Tan(0)| = 0.693-0 = 0.693
...........................
>> Is that right?

= ln|Sec(π/4)+Tan(π/4)| - ln|Sec(0)+Tan(0)| =
= ln|sqrt(2) + 1| - ln| 1 + 0| =

= ln|sqrt(2) + 1|

= 0.88137
..........................................
gives a different result from integrating sec(x) from -π/4 to 0  =
ln|Sec(0)+Tan(0)-ln|Sec(-π/4)+Tan(-π/4)| = 0-( -36.737)= 36.737)
.......................................
>> Wait a second --

ln|Sec(0)+Tan(0) - ln|Sec(-π/4)+Tan(-π/4)| =

ln| 1 + 0 | - ln| sqrt(2) - 1 |

- ln| sqrt(2) - 1 |

0.88137

>> So they are the same.  
As they should be.  Sec x is an even function, so the integral on either side of zero should be the same. (assuming you don't hit any singular points, like pi/2, etc.)