well, #sin x cos x = (sin 2x) /2# so you are looking at #1/2 int \ sin 2x \ dx = (1/2) [( 1/2) ( -cos 2x) + C] = -1/4 cos 2x + C'#
or maybe easier you can notice the pattern that # (sin^n x)' = n sin ^{n-1} x cos x# and pattern match. here #n-1 = 1# so n = 2 so we trial #(sin^2 x)'# which gives us # color{red}{2} sin x cos x# so we now that the anti deriv is #1/2 sin^2 x + C#
the other pattern also works ie # (cos^n x)' = n cos ^{n-1} x (-sin x) = - n cos ^{n-1} x sin x#
so trial solution #(-cos^2 x)' = -2 cos x (-sin x) = 2 cos x sin x# so the anti deriv is #-1/2 cos^2 x + C#
