Three different-looking answers to one integral

Three students find \(\displaystyle\int2\sin x\cos x\,\mathrm dx\) and obtain

\(\sin^2x+C,\qquad -\cos^2x+C,\qquad -\dfrac12\cos2x+C\).

1. Show, by differentiation, that each answer is correct.[3]
2. Show that any two of the three expressions \(\sin^2x\), \(-\cos^2x\) and \(-\dfrac12\cos2x\) differ by a constant, and explain why this means there is no contradiction.[3]
3. Show in the same way that \(\displaystyle\int\dfrac{1}{3x}\,\mathrm dx\) can be written as \(\dfrac13\ln x+C\) or as \(\dfrac13\ln(3x)+C\) for \(x>0\).[2]