How do you find the integral of #(sinx)(cosx)dx#?
There are several ways to write the correct answer.
#u = 2x#to get #1/4 cos(2x) + CC#
(I love this problem and use it every time I teach Calulus I.)
Note 1: It is not an accident that i used different notations for the constants in each solution.
The difference (literally -- that is, the subtraction) between the apparently different solutions are constants.