Evaluate the definite integral from pi/8 to 0 of sec2x tan2x

Question

Please show work!!!

Answer 1

∫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.