Problem with indefinite integral

Question

We have $\frac{d}{dx}\sin^2{x} = 2\sin{x}\cos{x} = \sin{2x}$. So what is wrong with $\int \sin{2x}dx =\int\frac{d}{dx} \sin^2{x}dx = \sin^2{x}?$ It cannot be correct since $\int \sin{2x}dx = -\frac{\cos{2x}}{2} \neq \sin^2{x}$.

Answer 1

This is where the importance of a constant of integration comes in. If you recall $\frac{1-\cos(2x)}{2}=\sin^2 x$ if you get two different representations for the same integral (assuming both are correct) then they only differ by a constant