Evaluate the integral (help)!! sin2x dx b=pi/8 a=0 ∫sin2x dx b=pi/8 a=0 I'm having a little trouble finishing out the problem. This is what i've got so far. u=2x du=2dx du= 1/2dx 1/2∫sin(u) du (-1/2)cos u (-1/2) cos 2x [(-1/2)cos2(pi/8)]-[(-1/2)cos(2(0))] ????? + (1/2) So... The homework question gave me the answer already, but I do not understand how the answer is .... (2 - sqrt(2) / 4) Can someone please "spell it out for me"... Thanks for your help!!