∫sec(2x)tan(2x)dx=½sec(2x). If we plug in the limits we get:
½sec(2π/8)=½sec(π/4)=1/(2cos(π/4))=1/√2=√2/2, because cos(π/4)=1/√2.
½sec(0)=½.
So the difference is ½(√2-1) if the high limit is π/8 and the low limit is 0;
or, -½(√2-1) if the high limit is 0 and the low limit is π/8.