Assume we have:
$ \int{ \cos{x} + \sin{x}\cos{x} dx } $
2 ways to do it:
Use $\sin{x}\cos{x} = \frac{ \sin{2x} }{2} $
Then $ \int{ \cos{x} + \frac{\sin{2x}}{2} dx } $
$ = \sin{x} - \frac{ cos{2x} }{ 4 } + C $
Or the other way, just see that $ u = \sin(x), du = \cos(x)dx $
$ \int{ \cos{x} + \sin{x}\cos{x} dx } $
$ = \sin{x} + \frac{\sin^2{x}}{2} + C $
Now the part I don't see fully is, why aren't these results completely equal?
Taking the 2nd result,
$ \sin{x} + \frac{\sin^2{x}}{2} $
$ = \sin{x} + \frac{1}{2} ( \frac{1 - cos{2x}}{2} ) $
$ = \sin{x} + \frac{1}{4} - \frac{\cos{2x}}{4} $
So you have to absorb $\frac{1}{4}$ into C for them to be equal.
Shouldn't they be equal straight away?