<s>[INST] 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.
 [/INST] $\displaystyle \int \sin x\cos xdx = \displaystyle \int \dfrac{\sin 2x}{2}dx = -\dfrac{\cos 2x}{4}+C$
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