Primitive of $\sin(4x)$ If $\sin(4x)=2\sin(2x)\cos(2x)$ then:
$$\int \sin(4x)=\int 2\sin(2x)\cos(2x)$$
$$\frac{-\cos(4x)}{4}=\frac{-\cos^2(2x)}{2}$$
But considering now $x=2$:
$$\frac{-\cos(4\cdot 2)}{4}\neq\frac{-\cos^2(2\cdot 2)}{2}$$
$$0.036\neq -0.214$$
What's the error that is behind this process ?
Thanks in advance.

A primitive is defined modulo a constant. And indeed a duplication formula yields
$$\cos 4x =2\cos^2 2x-1.$$
