-1/4 cos 2x + C
or 1/2 sin^2 x + C
or -1/2 cos^2 x + C
well, sinxcosx=sin2x2 so you are looking at12∫ sin2x dx=(12)[(12)(−cos2x)+C]=−14cos2x+C'
or maybe easier you can notice the pattern that (sinnx)'=nsinn−1xcosx and pattern match. here n−1=1 so n = 2 so we trial (sin2x)' which gives us 2sinxcosx so we now that the anti deriv is 12sin2x+C
the other pattern also works ie(cosnx)'=ncosn−1x(−sinx)=−ncosn−1xsinx
so trial solution (−cos2x)'=−2cosx(−sinx)=2cosxsinx so the anti deriv is −12cos2x+C
