what is the value of $\int \sin(x)\cos(x)dx$? $\frac{\sin^2(x)}{2}$ or $\frac{-\cos^2(x)}{2}$ or $\frac{-\cos(2x)}{4}$ $\int \sin(x)\cos(x)dx = \frac{\sin^2(x)}{2}$ because

$$\frac{d}{dx}\frac{\sin^2(x)}{2}=\sin(x)\frac{\sin(x)}{dx}=\sin(x)\cos(x)$$

but also

$$\frac{d}{dx}\frac{-\cos^2(x)}{2}=-\cos(x)\frac{\cos(x)}{dx}=\sin(x)\cos(x)$$

Moreover

$$\int \sin(x)\cos(x)dx = \int \frac{\sin(2x)}{2}dx = \frac{-\cos(2x)}{4} = \frac14 -  \frac{\cos^2(x)}{2} = \frac{\sin^2(x)}{2} - \frac14 $$

It's true that $\frac{\sin^2(x)}{2}$ and $\frac{-\cos^2(x)}{2}$ are not equal so what is that problem I made here.
You are forgetting the constant of integration.  The constant of integration will change depending on whether you chose to integrate the sine or the cosine.  The bottom line is that the integrals are the same.
$\displaystyle \int \sin x\cos xdx = \displaystyle \int \dfrac{\sin 2x}{2}dx = -\dfrac{\cos 2x}{4}+C$